Alternative solution continuing from 0:54, sqrt(ab) is an integer, ab is a perfect square. Then a = k*n^2 and b = k*m^2, where k,m,n are nonnegative integers and no factor of k is a perfect square other than 1. Put back into original equation to have (m+n)*sqrt(k) = sqrt(2009) k*(m+n)^2 = 41*7^2 m+n = 7, k = 41 (m,n)=(0,7),(1,6)...(6,1),(7,0) (a,b)=(0,2009),(41,1376)...(1476,41),(2009,0)
@@AlephThree “ab is a perfect square, so a=k*m^2 and b=k*n^2, where k is a positive integer without any factor being a square other than 1, and m and n are nonnegative.” In your example 5 = 5*1^2 while 20 = 5*2^2
I solved it in very short and simple way : Sqrt A + Sqrt B = 7sqrt 41 => Sqrt A/41 + sqrt B/41 = 7. Since 7 is integer therefore a = 41k^2 and b=41m^2 => k + m = 7. Hence k,m = (0,7) (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) and (7,0) which brings us to the final answer.
Nice solution. When you had b=41k^2, you also have by symmetry that a=41m^2 for some non negative integer m. Then if you go back to the original equation and substitute in, this reduces down to k+m=7. Then the pairs (k,m) are (0,7), (1,6),...(7,0). Just a slightly different way of doing this...
Alternative solution: 2009 can be written as 7² * 41 thus √(2009) = 7√(41) So if you multiply both sides of the equation by √(41) RHS becomes integer. Thus LHS should become an integer as well. √(41)(√a + √b) = 7*41 --- (1) √(41)(√a + √b) is integer. So we can say a=41x² ; b=41y² By substitution to (1) equation we get 41x + 41y = 7*41 x+y=7 Now it is easy to find all the positive integer solutions for x and y and then a and b
Interesting problem and nice solution. Alternative approach: √a + √b = √2009 = 7√41. Solution is (√a, √b) = (m√41, (7-m)√41), i.e. (a,b) = (41m², 41(7-m)²), for m = 0,1,2,..,7.
@@mitjordan You can think of it as equating coefficients for the radical part. See en.wikipedia.org/wiki/Equating_coefficients#Example_in_nested_radicals.
I did it in a very easy way. After squaring, I divided by a to get 1+b/a+2√(b/a) = 2009 Since RHS is rational, so must the LHS be - this means √(b/a) is rational. Now put √(b/a) as x. You get a quadratic equation x^2 + 2x + (1 - 2009/a) = 0 The solution to this is -1+√(2009/a) Now this solution must be rational, as mentioned earlier. So, 2009/a must be a perfect square k^2, and must be rational, so k is also rational So a = 2009/k^2 => a = 41x49/k^2 For a to be an integer, k^2 must cancel out with the numerator. It can't contain 41 as that is not a square. So it must contain 49 So, k = 7/n => a = 41 x n^2 Also, k is a square root (√2009/a) it must be > 0. So 7/n >0 so n>0. Furthermore, √(b/a) was equal to -1+√(2009/a). This must be not only rational, but non negative too, since it is a square root. Which means that √(2009/a) >= 1. Put a = 41 x n^2. This leads to n
sqrt(a)+sqrt(b)=sqrt(2009) Squaring --> a+b+2sqrt(ab)=2009 Dividing by a we get 1+(b/a)+2sqrt(b/a)=2009/a not 1+(b/a)+2sqrt(b/a)=2009 in the end of first sentence of your comment.
Another solution. Note (√a + √b)(√a - √b) = a - b is an integer k. Thus, we deduce that √a - √b = k/√2009. Now we can solve for √a,√b in expressions in k. Multiply these expressions to get √ab in terms of k. From 0:54, ab must be a square number. The result follows by exhaustion.
Whenever I see an integer which seems to have no relation to anything, I feel the urge to see if it has any factors. As you discovered 2009 = 49*41 that is a square multiplied by a prime. So we can write the original equation as sqrt(a) + sqrt(b) = 7*sqrt(41) If we put a = m^2*41 and b = n^2*41 the original equation becomes sqrt(m^2*41) = sqrt(n^2*41) = 7*sqrt(41) -> m + n := 7 So the solution set now is easier to see.
You can't just "put" a=41*m^2... For example, maybe there are some a,b,c s.t. a*sqrt(10)+b*sqrt(11)=c*sqrt(41). You must prove that a and b are 41*m^2, respectively 41*n^2.
Once you get to sqrt(a) + sqrt(b) = 7*sqrt(41), it's easy to see sqrt(a) + sqrt(b) = sqrt(41)+6sqrt(41) sqrt(a) + sqrt(b) = 2sqrt(41)+5sqrt(41) sqrt(a) + sqrt(b) = 3sqrt(41)+4sqrt(41) and so on, so the set of solutions for a and b are easy to calculate. There can't be any other whole number solutions because 2009 has no other whole number factors which are also squares.
But "4ab is a perfect square" *is* helpful. It means ab is a perfect square, hence a = u²w and b = v²w for non-negative intergers u,v,w. So sqrt(a)+sqrt(b) = (u+v)sqrt(w) and we are practically done, namely w must be a divisor of 2009 with square cofactor
continuing from where he left: (u+v)sqrt(w) = sqrt(2009) (.)² (u+v)²w= 2009 ; this is a diophantine eqn. 2009= 2009×1 ; 7×287 , 49×41 only pair that satisfies is 49×41 Thus (u+v)²=49 --> u+v= 7 Therefore 8 solutions
bro great video. and thing i liked the most is that u showed what is wrong first. that thing is very helpful for someone seeing a new kind of problem. thank you and pls keep doing that :)
We may also write √a=7√41-√b. We obtain 41b is an integer, so b=41b'. And by doing the same for a: a=41a'. The equation becomes √a'+√b'=7. By the same method a' is a square between 0 and 49.
When you suggest that b=41k^2 and you plug it back to the original formula (timecode 4:45), it seems that you do a mistake. Under the radical in the original equation, you have only 2009b. So when you plug b here, you should get 7^2*41*41k^2 - but in your solution (timecode 5:00) you also have 2^2 under the square root.
What i did was i found root 2009 as 7root 41, so the only thing what can make root 2009 is 1root41+6root41, 2root41+5root41, 3root41+4root41 and root0 + root 2009, from thereci sqaured those numbers and found the integers that way
Use the sum formula 1/40 + 2/40 + 3/40....+39/40 Rewrite as (n(n+1))/40 Let n be 39 because we went from 1 to 39 Rewrite ((39(39+1))/40) -(1/2) (780/40)-(1/2)= 19 Helps to know the sum of the natural numbers as well as squares and cubes.
Beloved, Please give this your special attention. God is a Good, Holy and Just( meaning that He up holds justice and does what He deems as good) Judge. He is also forgiving and He is All-Loving. He is also perfect meaning that there is nothing about Him this is contradictory. But How can He forgive you and me but still uphold Justice. Look at it in this way: Judges who punish the innocent and release the wicked and quilty, are wicked, evil and are corrupt. It does not matter how much a quilty man is sorry for his crime, he must go to jail. Let's say a man raped and murdered a little girl, but is so sorry for the rape; the judge must still send him to jail. And likewise God must still punish our sins, but He can still forgive us. How? How can God forgive us and still satisfy His Justice? This is how: All have sinned and fall short of the glory of God. There is none righteous, no not one” (Romans 3:10, 23 NKJV). -It means that each and every single person sins and has sinned. “The wages of sin is death, but the gift of God is eternal life through Jesus Christ our Lord” (Romans 6:23). -Death is eternal separation from God in hell. - But believing in Christ alone allows you to be with God in everlasting joy. •"The iniquity and sins of each and every persons were placed upon Jesus Christ who was sinless. •Jesus Christ, the Son of God was crucified for our sins. •Jesus Christ was buried and our sins were buried with Him to live no more. •Jesus Christ arose from the dead on the third day and all who believe and receive Him as Lord arose to new life with Him. • New life in Jesus Christ is received through faith in His death burial and resurrection, when the person repents of their sin and asks Jesus Christ to forgive their sin and wash their heart. Beloved seek Him, read the Bible, not the catholic Bible(deutero-canonical books) Start by read the whole of the new testament, starting from Matthew to Revelation.. i love you because Christ loves me.
let a = m*sqrt(x) b = n*sqrt(y), where m,n are integers, and x, y no longer have square number as factors. assume x=q1*q2*...*qk, and y has to be dividable for each qi, so y has to equal to x. so it is (m+n)sqrt(x)=7*sqrt(41), so m = 0,1,2...7 n=7,6,5...0 correspondingly
Otherwise, there is a famous Lemma which says that if gcd.(a,b)=d and ab is a perfect square, then a=dx², b=dy², with gcd.(x,y)=1. Hence, via squaring, we have the equation: dx²+2dxy+dy²=2009 d(x+y)²=2009=41·7². So d=41 and x+y=±7.
I think you did this the long way around. I started by simplifying root2009 into 7root41. Then we can have pairs for a and b root41, 6root41; 2root41, 5root41; 3root 41, 4root41; and their reversals. There's no other ways to divide that up. (And the 0, 7root41 pair of pairs - thought it was positive, not non-negative.)
@@MrAmangandhi Actually this can be proven with my method, I think. Basically it's the only method to divide this number into additive pairs. root41 cannot be divided by further to yield an integer. I don't know the formal way to present it.
@@ryanjagpal9457 It was pretty easy to find that sqrt2009 = 7 * sqrt41. From there I guess a and b must be (1*sqrt41, 6*sqrt41), (2*sqrt41, 5*sqrt41)…(6*sqrt41, 1*sqrt41). But like I said, I don’t have a proof that those are the only answers.
@@ryanjagpal9457 And no, I didn’t memorize 2009 is a multiply of 49. But I knew it’s not a multiply of 2,3,or 5. And it just happened to be a multiply of 7, and 287 also happened to be a multiply of 7
2009=7²41 √a=√(2009) - √b a=2009+b-2√(2009b) but a is an integer so √(2009b) = 7√(41b) is an integer so b=y²41, similarly (symmetry) a=x²41 original eqn => x√41 + y√41 =7√41 x + y = 7 so x runs from 0 to 1 as y runs from 7 to zero thus solutions (of the form n²41) are ... 0,2009 41,1476 164,1025 369,656 656,369 1025,164 1476,41 2009,0
For someone who has a horrible maths education, I find these solutions so much cleaner and easier to understand than Micheal Penn’s as I am familiar with most of the terms involved here. Whereas I don’t know what fermats little theorem is
I don’t understand these random things like calculus, it’s stupid and isn’t logic like (a+b)(a-b) is a^2 - b^2 And it makes no sense how people understand this sort of stuff, are they smart?
@@ryanjagpal9457 I’d say they’re smart if they compete in olympiads. However, anyone can understand this stuff if they spend enough time at it. Adults who don’t have a natural talent for maths, but maybe spent their time studying maths at uni and afterwards would probably find this much easier than a 15 yo maths genius. Being bad at this doesn’t mean you’re bad at maths, just that you haven’t spent enough time at it. I sound so wise lol. But I’m talking from experience. I’m always horrible at stuff I start but then they become super easy after a while.
@@NegativeAccelerate Yeah, sometimes I learn trigonometry online lessons and i didn’t understand it, and the back to school i understand it face to face I am pretty good at maths, I’m not on that calculus level, no god no
Recently a co-worker presented same problem, but with current year 2023 in the rhs sqrt. 2023 is also in the form p*q^2 with p, q primes, so it can be solved in same way. With p,q primes, next time this can be an IMO problem is 2043.
This is how I did it :- √a+√b=√2009 √a+√b=7√41 Now, a and b are integers. So, a and b must be multiple of 41 (and not of 41^2) such that √a+√b=7√41 So, the possible values of (a,b) are (41*0^2, 41*7^2) (41*1^2, 41*6^2) (41*2^2, 41*5^2) (41*3^2, 41*4^2) (41*4^2, 41*3^2) (41*5^2, 41*2^2) (41*6^2, 41*1^2) (41*0^2, 41*7^2) There are 8 possible values of a and b. I am still a high schooler, and learning, so let me know if i have overlooked something here
i squared equation and got a+2sqrt(ab)+b=2009 which implies ab=k^2 for some nonegative integer k (ignoring zero solutions). then this let's us solve equation which via quadratic formula yields k=-b+sqrt(b*2009) hence implying b=41*l^2 for some positive integer l. plugging this into ab=k^2 returns a=41(7-l)^2 which is same as video. (also 0
Although directly factorizing 2009 is the most elegant solution, I want to share an analytic solution which requires less “magic” as presented in the video. The key is the odd-even parity on both sides. First, square both sides: a+b+2sqrt(ab) = 2009 - eq(1) We can see that the right hand side is ODD. So a, b must be one ODD and one EVEN. We can assume a is ODD and b is EVEN. Because we have to ensure that ab is a square number. Therefore, we can let b = 4pq^2 a = pr^2 (So that ab = (2pqr)^2) ,where p,q,r are some positive integers. Plugging in eq(1), we have p(r+2q)^2 = 2009 So we know that, p = 41 and r+2q = 7, so that three sets of solutions are: (p,q,r) = (41,1,5), (41,2,3),(41,3,1) Remember that we have assumed a is ODD and b is EVEN, you can also have b ODD and a EVEN. (Of course there is also two trivial solutions.)
a+b+2root(ab)=2009, product ab has to be perfect square,a=b or 4b or 9b or 36b.... 1024b, prime factors of 2009 are 1, 7, 41, 49, 287 and 2009, at 36b solution is, 36b+b+2root(36b2)=>49b=2009,a=1476,b=41.
Plugging the b=41k^2 back and evaluating a to put upper limit on k was unnecessary. We know sqrt(a) and sqrt(b) are both less or equal to sqrt(2009), So, a and be are both less or equal to 2009. 2009 is 41*7^2 which means that k^2 less or equal to 7^2 so k is less or equal to 7.
B cannot be equal to a due to them being integral Hence root b less than root a 2 root b less than root 2009 B less than 2009 / 4 Now u do the thing done in the video and u get a = 2009 + b - 2 root 2009×b Hence for (a) to be integer b = 41 k sq 41k sq less than 2009/4 Hence k = 0 1 2 3 So here we get all solutions in which a greater than b Same we do for b greater than a and get all the 8 solutions
So he squares both sides, the first side is basically sqrt(a)+sqrt(b) (sqrt=square root), in order to square that you write it like this (sqrt(a)+sqrt(b))^2 and that is equal to sqrt(a)^2+2sqrt(ab)+sqrt(b)^2 (check binomial theorem) so the squares cancel the roots and you end up with a+2sqrt(ab)+b and the other side it's just square root of 2009 to the power of 2 and you end with just 2009
Beloved, Please give this your special attention. God is a Good, Holy and Just( meaning that He up holds justice and does what He deems as good) Judge. He is also forgiving and He is All-Loving. He is also perfect meaning that there is nothing about Him this is contradictory. But How can He forgive you and me but still uphold Justice. Look at it in this way: Judges who punish the innocent and release the wicked and quilty, are wicked, evil and are corrupt. It does not matter how much a quilty man is sorry for his crime, he must go to jail. Let's say a man raped and murdered a little girl, but is so sorry for the rape; the judge must still send him to jail. And likewise God must still punish our sins, but He can still forgive us. How? How can God forgive us and still satisfy His Justice? This is how: All have sinned and fall short of the glory of God. There is none righteous, no not one” (Romans 3:10, 23 NKJV). -It means that each and every single person sins and has sinned. “The wages of sin is death, but the gift of God is eternal life through Jesus Christ our Lord” (Romans 6:23). -Death is eternal separation from God in hell. - But believing in Christ alone allows you to be with God in everlasting joy. •"The iniquity and sins of each and every persons were placed upon Jesus Christ who was sinless. •Jesus Christ, the Son of God was crucified for our sins. •Jesus Christ was buried and our sins were buried with Him to live no more. •Jesus Christ arose from the dead on the third day and all who believe and receive Him as Lord arose to new life with Him. • New life in Jesus Christ is received through faith in His death burial and resurrection, when the person repents of their sin and asks Jesus Christ to forgive their sin and wash their heart. Beloved seek Him, read the Bible, not the catholic Bible(deutero-canonical books) Start by read the whole of the new testament, starting from Matthew to Revelation.. i love you because Christ loves me.
@@average_loser4254 Beloved, Please give this your special attention. God is a Good, Holy and Just( meaning that He up holds justice and does what He deems as good) Judge. He is also forgiving and He is All-Loving. He is also perfect meaning that there is nothing about Him this is contradictory. But How can He forgive you and me but still uphold Justice. Look at it in this way: Judges who punish the innocent and release the wicked and quilty, are wicked, evil and are corrupt. It does not matter how much a quilty man is sorry for his crime, he must go to jail. Let's say a man raped and murdered a little girl, but is so sorry for the rape; the judge must still send him to jail. And likewise God must still punish our sins, but He can still forgive us. How? How can God forgive us and still satisfy His Justice? This is how: All have sinned and fall short of the glory of God. There is none righteous, no not one” (Romans 3:10, 23 NKJV). -It means that each and every single person sins and has sinned. “The wages of sin is death, but the gift of God is eternal life through Jesus Christ our Lord” (Romans 6:23). -Death is eternal separation from God in hell. - But believing in Christ alone allows you to be with God in everlasting joy. •"The iniquity and sins of each and every persons were placed upon Jesus Christ who was sinless. •Jesus Christ, the Son of God was crucified for our sins. •Jesus Christ was buried and our sins were buried with Him to live no more. •Jesus Christ arose from the dead on the third day and all who believe and receive Him as Lord arose to new life with Him. • New life in Jesus Christ is received through faith in His death burial and resurrection, when the person repents of their sin and asks Jesus Christ to forgive their sin and wash their heart. Beloved seek Him, read the Bible, not the catholic Bible(deutero-canonical books) Start by read the whole of the new testament, starting from Matthew to Revelation.. i love you because Christ loves me.
I usually enjoy the way you approach Math problems. However your approach here was tedious, in my humble opinion. I factorized 2009 as 7**2, 41. So, √a, √b are of the form p√41, q√41, where p+q = 7. Now read off all eight pairs.
@@TM-ht8jv Yes, you are right. Thank you for pointing this out. In order to help my argument, I must use sqrt(a) + sqrt(b) = m sqrt(r) where r is a prime number and m is an integer. Now dividing both sides by sqrt(r) and squaring we get: m**2 = (a+b)/r + 2 sqrt(ab)/r. Now I think we are in similar territory as the solution supplied by @letsthinkcritically.
Why have we ignored the possibility that the square roots can yield negative results too? In that case, the solution is (a,b) = (41*m^2, 41*n^2) where m+n=7 for all m,n in Integer space.
Alternative solution continuing from 0:54,
sqrt(ab) is an integer, ab is a perfect square.
Then a = k*n^2 and b = k*m^2, where k,m,n are nonnegative integers and no factor of k is a perfect square other than 1.
Put back into original equation to have (m+n)*sqrt(k) = sqrt(2009)
k*(m+n)^2 = 41*7^2
m+n = 7, k = 41
(m,n)=(0,7),(1,6)...(6,1),(7,0)
(a,b)=(0,2009),(41,1376)...(1476,41),(2009,0)
The same way as I did
If ab is a perfect square why does this make both a and b perfect squares? Eg 100 = 5 x 20, and neither 5 nor 20 are perfect squares.
@@AlephThree “ab is a perfect square, so a=k*m^2 and b=k*n^2, where k is a positive integer without any factor being a square other than 1, and m and n are nonnegative.”
In your example 5 = 5*1^2 while 20 = 5*2^2
@@nickcheng2547 thanks, I understand now
Hi, why is it k*m^2 and k*n^2? What if it they didn’t have a common factor, K?
I solved it in very short and simple way : Sqrt A + Sqrt B = 7sqrt 41 => Sqrt A/41 + sqrt B/41 = 7. Since 7 is integer therefore a = 41k^2 and b=41m^2 => k + m = 7. Hence k,m = (0,7) (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) and (7,0) which brings us to the final answer.
Best approach
@@ironbutterfly3701it's a well known result that if sqrt(x) + sqrt(y) is an integer then both x and y are perfect squares
@@ironbutterfly3701 Becuase sum of two positive irrational numbers can never be a rational number.
@@ravirajamadan That is not true, (5+sqrt(2)) + (3-sqrt(2)) = 8
@@AnitaSV Bur an irrational no. as shown by you cannot be a square root of an integer.
Nice solution. When you had b=41k^2, you also have by symmetry that a=41m^2 for some non negative integer m. Then if you go back to the original equation and substitute in, this reduces down to k+m=7. Then the pairs (k,m) are (0,7), (1,6),...(7,0). Just a slightly different way of doing this...
Wtf?
You're right. The right side has rt41 and this must be present to both a and b. Thus, it exactly implies your supposed equation is proven correct.
Alternative solution:
2009 can be written as 7² * 41
thus √(2009) = 7√(41)
So if you multiply both sides of the equation by √(41) RHS becomes integer. Thus LHS should become an integer as well.
√(41)(√a + √b) = 7*41 --- (1)
√(41)(√a + √b) is integer.
So we can say a=41x² ; b=41y²
By substitution to (1) equation we get
41x + 41y = 7*41
x+y=7
Now it is easy to find all the positive integer solutions for x and y and then a and b
Interesting problem and nice solution.
Alternative approach: √a + √b = √2009 = 7√41.
Solution is (√a, √b) = (m√41, (7-m)√41),
i.e. (a,b) = (41m², 41(7-m)²), for m = 0,1,2,..,7.
I thought the same too - my only thought is that as "obvious" those are the 8 solutions, it is not a proof that there are no other solutions.
@@mitjordan You can think of it as equating coefficients for the radical part. See en.wikipedia.org/wiki/Equating_coefficients#Example_in_nested_radicals.
what are you saying, this made no sense
@@oxbmaths I think you are right BTW, but to get full points you would have to prove that root(41) is irrational.
@@mitjordan Thanks. We know that 41 is prime, so the root of 41 must be irrational.
I did it in a very easy way. After squaring, I divided by a to get 1+b/a+2√(b/a) = 2009
Since RHS is rational, so must the LHS be - this means √(b/a) is rational.
Now put √(b/a) as x. You get a quadratic equation x^2 + 2x + (1 - 2009/a) = 0
The solution to this is -1+√(2009/a)
Now this solution must be rational, as mentioned earlier. So, 2009/a must be a perfect square k^2, and must be rational, so k is also rational
So a = 2009/k^2
=> a = 41x49/k^2
For a to be an integer, k^2 must cancel out with the numerator. It can't contain 41 as that is not a square. So it must contain 49
So, k = 7/n
=> a = 41 x n^2
Also, k is a square root (√2009/a) it must be > 0. So 7/n >0 so n>0.
Furthermore, √(b/a) was equal to -1+√(2009/a). This must be not only rational, but non negative too, since it is a square root. Which means that √(2009/a) >= 1. Put a = 41 x n^2. This leads to n
sqrt(a)+sqrt(b)=sqrt(2009)
Squaring --> a+b+2sqrt(ab)=2009
Dividing by a we get
1+(b/a)+2sqrt(b/a)=2009/a
not 1+(b/a)+2sqrt(b/a)=2009
in the end of first sentence of your comment.
Another solution. Note (√a + √b)(√a - √b) = a - b is an integer k. Thus, we deduce that √a - √b = k/√2009. Now we can solve for √a,√b in expressions in k. Multiply these expressions to get √ab in terms of k. From 0:54, ab must be a square number. The result follows by exhaustion.
Whenever I see an integer which seems to have no relation to anything, I feel the urge to see if it has any factors. As you discovered 2009 = 49*41 that is a square multiplied by a prime.
So we can write the original equation as
sqrt(a) + sqrt(b) = 7*sqrt(41)
If we put a = m^2*41 and b = n^2*41 the original equation becomes
sqrt(m^2*41) = sqrt(n^2*41) = 7*sqrt(41)
-> m + n := 7
So the solution set now is easier to see.
Yes. It is straight forward and simpler.
You first need to discover that 41 | a and 41 | b.
You can't just "put" a=41*m^2...
For example, maybe there are some a,b,c s.t. a*sqrt(10)+b*sqrt(11)=c*sqrt(41).
You must prove that a and b are 41*m^2, respectively 41*n^2.
what if a and b are not in that form? dont you have to prove those aren't solutions
Once you get to sqrt(a) + sqrt(b) = 7*sqrt(41), it's easy to see
sqrt(a) + sqrt(b) = sqrt(41)+6sqrt(41)
sqrt(a) + sqrt(b) = 2sqrt(41)+5sqrt(41)
sqrt(a) + sqrt(b) = 3sqrt(41)+4sqrt(41)
and so on,
so the set of solutions for a and b are easy to calculate.
There can't be any other whole number solutions because 2009 has no other whole number factors which are also squares.
Its not done until you say " and yay we are done =) "
Easy to see 2009*a and 2009*b are perfect squares, so a = 41m^2 and b = 41n^2.
Substitute to get m+n = 7 and all 8 solutions follows.
Why 7?
@@ryanjagpal9457 sqrt(2009) = 7.sqrt(41)
@@heldercomp Ok sir
7:12 Why do we take (7-k) instead of (k-7)
But "4ab is a perfect square" *is* helpful. It means ab is a perfect square, hence a = u²w and b = v²w for non-negative intergers u,v,w. So sqrt(a)+sqrt(b) = (u+v)sqrt(w) and we are practically done, namely w must be a divisor of 2009 with square cofactor
What you saying cause I don’t understand?
continuing from where he left:
(u+v)sqrt(w) = sqrt(2009)
(.)²
(u+v)²w= 2009 ; this is a diophantine eqn.
2009= 2009×1 ; 7×287 , 49×41
only pair that satisfies is 49×41
Thus (u+v)²=49 --> u+v= 7
Therefore 8 solutions
Two oh oh nine!
Never heard that before.
And a great solution. Nice
bro great video. and thing i liked the most is that u showed what is wrong first. that thing is very helpful for someone seeing a new kind of problem. thank you and pls keep doing that :)
We may also write √a=7√41-√b. We obtain 41b is an integer, so b=41b'.
And by doing the same for a: a=41a'.
The equation becomes √a'+√b'=7.
By the same method a' is a square between 0 and 49.
When you suggest that b=41k^2 and you plug it back to the original formula (timecode 4:45), it seems that you do a mistake. Under the radical in the original equation, you have only 2009b. So when you plug b here, you should get 7^2*41*41k^2 - but in your solution (timecode 5:00) you also have 2^2 under the square root.
sorry, just watched further and see you corrected it.
I’m confused even if you were right
What i did was i found root 2009 as 7root 41, so the only thing what can make root 2009 is 1root41+6root41, 2root41+5root41, 3root41+4root41 and root0 + root 2009, from thereci sqaured those numbers and found the integers that way
Bro, you cooked
Use the sum formula
1/40 + 2/40 + 3/40....+39/40
Rewrite as (n(n+1))/40
Let n be 39 because we went from 1 to 39
Rewrite ((39(39+1))/40) -(1/2)
(780/40)-(1/2)= 19
Helps to know the sum of the natural numbers as well as squares and cubes.
Really nicely done. Liked and subbed.
Beloved, Please give this your special attention.
God is a Good, Holy and Just( meaning that He up holds justice and does what He deems as good) Judge. He is also forgiving and He is All-Loving. He is also perfect meaning that there is nothing about Him this is contradictory. But How can He forgive you and me but still uphold Justice.
Look at it in this way: Judges who punish the innocent and release the wicked and quilty, are wicked, evil and are corrupt. It does not matter how much a quilty man is sorry for his crime, he must go to jail. Let's say a man raped and murdered a little girl, but is so sorry for the rape; the judge must still send him to jail. And likewise God must still punish our sins, but He can still forgive us. How? How can God forgive us and still satisfy His Justice?
This is how:
All have sinned and fall short of the glory of God. There is none righteous, no not one” (Romans 3:10, 23 NKJV).
-It means that each and every single person sins and has sinned.
“The wages of sin is death, but the gift of God is eternal life through Jesus Christ our Lord” (Romans 6:23).
-Death is eternal separation from God in hell.
- But believing in Christ alone allows you to be with God in everlasting joy.
•"The iniquity and sins of each and every persons were placed upon Jesus Christ who was sinless.
•Jesus Christ, the Son of God was crucified for our sins.
•Jesus Christ was buried and our sins were buried with Him to live no more.
•Jesus Christ arose from the dead on the third day and all who believe and receive Him as Lord arose to new life with Him.
• New life in Jesus Christ is received through faith in His death burial and resurrection, when the person repents of their sin and asks Jesus Christ to forgive their sin and wash their heart.
Beloved seek Him, read the Bible, not the catholic Bible(deutero-canonical books)
Start by read the whole of the new testament, starting from Matthew to Revelation..
i love you because Christ loves me.
Very good explanation & technique. Thanks.
Thank you Sir. Brilliantly solved.
let a = m*sqrt(x) b = n*sqrt(y), where m,n are integers, and x, y no longer have square number as factors. assume x=q1*q2*...*qk, and y has to be dividable for each qi, so y has to equal to x. so it is (m+n)sqrt(x)=7*sqrt(41), so m = 0,1,2...7 n=7,6,5...0 correspondingly
Otherwise, there is a famous Lemma which says that if gcd.(a,b)=d and
ab is a perfect square, then
a=dx², b=dy², with gcd.(x,y)=1. Hence, via squaring, we have the equation:
dx²+2dxy+dy²=2009
d(x+y)²=2009=41·7².
So d=41 and x+y=±7.
Nice lesson! :D
I think you did this the long way around. I started by simplifying root2009 into 7root41. Then we can have pairs for a and b root41, 6root41; 2root41, 5root41; 3root 41, 4root41; and their reversals. There's no other ways to divide that up. (And the 0, 7root41 pair of pairs - thought it was positive, not non-negative.)
thats true...but his approach also proves why they are only solutions
@@MrAmangandhi Actually this can be proven with my method, I think. Basically it's the only method to divide this number into additive pairs. root41 cannot be divided by further to yield an integer. I don't know the formal way to present it.
Confused
can you please include problems from combinatorial topics like graph theory oriented in a fashion for olympiads. please?
From where did you study number theory?
What is that?
I got the same result in less than a min in my head, but couldn't prove if there aren't any other pairs in my head.
So you guessed it? Probably the same
@@ryanjagpal9457 It was pretty easy to find that sqrt2009 = 7 * sqrt41. From there I guess a and b must be (1*sqrt41, 6*sqrt41), (2*sqrt41, 5*sqrt41)…(6*sqrt41, 1*sqrt41). But like I said, I don’t have a proof that those are the only answers.
@@ryanjagpal9457 And no, I didn’t memorize 2009 is a multiply of 49. But I knew it’s not a multiply of 2,3,or 5. And it just happened to be a multiply of 7, and 287 also happened to be a multiply of 7
@@Ranoiaetep 1, 2, 6, 6 5, 1
It’s way easier if you find multiples
2009=7²41
√a=√(2009) - √b
a=2009+b-2√(2009b) but a is an integer so √(2009b) = 7√(41b) is an integer
so b=y²41,
similarly (symmetry)
a=x²41
original eqn =>
x√41 + y√41 =7√41
x + y = 7
so x runs from 0 to 1 as y runs from 7 to zero
thus solutions (of the form n²41) are ...
0,2009 41,1476 164,1025 369,656 656,369 1025,164 1476,41 2009,0
For someone who has a horrible maths education, I find these solutions so much cleaner and easier to understand than Micheal Penn’s as I am familiar with most of the terms involved here. Whereas I don’t know what fermats little theorem is
I don’t understand these random things like calculus, it’s stupid and isn’t logic like (a+b)(a-b) is a^2 - b^2
And it makes no sense how people understand this sort of stuff, are they smart?
@@ryanjagpal9457 I’d say they’re smart if they compete in olympiads. However, anyone can understand this stuff if they spend enough time at it.
Adults who don’t have a natural talent for maths, but maybe spent their time studying maths at uni and afterwards would probably find this much easier than a 15 yo maths genius.
Being bad at this doesn’t mean you’re bad at maths, just that you haven’t spent enough time at it.
I sound so wise lol. But I’m talking from experience. I’m always horrible at stuff I start but then they become super easy after a while.
@@NegativeAccelerate Yeah, sometimes I learn trigonometry online lessons and i didn’t understand it, and the back to school i understand it face to face
I am pretty good at maths, I’m not on that calculus level, no god no
Recently a co-worker presented same problem, but with current year 2023 in the rhs sqrt. 2023 is also in the form p*q^2 with p, q primes, so it can be solved in same way. With p,q primes, next time this can be an IMO problem is 2043.
This is how I did it :-
√a+√b=√2009
√a+√b=7√41
Now, a and b are integers. So, a and b must be multiple of 41 (and not of 41^2) such that √a+√b=7√41
So, the possible values of (a,b) are
(41*0^2, 41*7^2)
(41*1^2, 41*6^2)
(41*2^2, 41*5^2)
(41*3^2, 41*4^2)
(41*4^2, 41*3^2)
(41*5^2, 41*2^2)
(41*6^2, 41*1^2)
(41*0^2, 41*7^2)
There are 8 possible values of a and b.
I am still a high schooler, and learning, so let me know if i have overlooked something here
i squared equation and got a+2sqrt(ab)+b=2009 which implies ab=k^2 for some nonegative integer k (ignoring zero solutions). then this let's us solve equation which via quadratic formula yields k=-b+sqrt(b*2009) hence implying b=41*l^2 for some positive integer l. plugging this into ab=k^2 returns a=41(7-l)^2 which is same as video. (also 0
You should use symmetry. He showed b=41k^2 so with the same reasoning, a=41^2 t. Now, you end up getting k+t=7 and there are 8 solutions.
(a, b) = {0, 2009} , {41, 1476} , {164, 1025} , {369, 656} .
I think this is much easier if you just factor 2009 no?
Although directly factorizing 2009 is the most elegant solution, I want to share an analytic solution which requires less “magic” as presented in the video. The key is the odd-even parity on both sides.
First, square both sides: a+b+2sqrt(ab) = 2009 - eq(1)
We can see that the right hand side is ODD. So a, b must be one ODD and one EVEN. We can assume a is ODD and b is EVEN.
Because we have to ensure that ab is a square number. Therefore, we can let
b = 4pq^2
a = pr^2
(So that ab = (2pqr)^2)
,where p,q,r are some positive integers. Plugging in eq(1), we have
p(r+2q)^2 = 2009
So we know that, p = 41 and r+2q = 7, so that three sets of solutions are:
(p,q,r) = (41,1,5), (41,2,3),(41,3,1)
Remember that we have assumed a is ODD and b is EVEN, you can also have b ODD and a EVEN. (Of course there is also two trivial solutions.)
a+b+2root(ab)=2009, product ab has to be perfect square,a=b or 4b or 9b or 36b.... 1024b, prime factors of 2009 are 1, 7, 41, 49, 287 and 2009, at 36b solution is, 36b+b+2root(36b2)=>49b=2009,a=1476,b=41.
That reminds me of HKAL Pure Math or HKCEE A.Math.
Plugging the b=41k^2 back and evaluating a to put upper limit on k was unnecessary. We know sqrt(a) and sqrt(b) are both less or equal to sqrt(2009), So, a and be are both less or equal to 2009. 2009 is 41*7^2 which means that k^2 less or equal to 7^2 so k is less or equal to 7.
I mean it has to be equal to something
5:10 Why is it okay to ignore the negative value of the root of the green part? Is it something obvious?
there is no negative value what are you talking about ?
can anyone tell me the number of soutions to (abc)/(a+b+c)
*FindInstance[
Sqrt[a] + Sqrt[b] == Sqrt[2009] \[And] a >= 0 \[And] b >= 0, {a, b}, Integers, 100000]*
Thanks bro
8 no of solutions?
i solved the question it was the most beautiful question i have seen
u should try giving solutions of rmo inmo india
2009= 49x41
From this idea, we can obviously see the answer
B cannot be equal to a due to them being integral
Hence root b less than root a
2 root b less than root 2009
B less than 2009 / 4
Now u do the thing done in the video and u get
a = 2009 + b - 2 root 2009×b
Hence for (a) to be integer b = 41 k sq
41k sq less than 2009/4
Hence k = 0 1 2 3
So here we get all solutions in which a greater than b
Same we do for b greater than a and get all the 8 solutions
I’m confused with after b=41k^2
Thanks
I thint there are just 6 solutions because a and b are nonnegative so the couples (0, 41×(7^2)) and (41×(7^2) , 0) aren't solutions
The first step. Why do you do that? I want to end my life tbh. Im struggling.
So he squares both sides, the first side is basically sqrt(a)+sqrt(b) (sqrt=square root), in order to square that you write it like this (sqrt(a)+sqrt(b))^2 and that is equal to sqrt(a)^2+2sqrt(ab)+sqrt(b)^2 (check binomial theorem) so the squares cancel the roots and you end up with a+2sqrt(ab)+b and the other side it's just square root of 2009 to the power of 2 and you end with just 2009
Beloved, Please give this your special attention.
God is a Good, Holy and Just( meaning that He up holds justice and does what He deems as good) Judge. He is also forgiving and He is All-Loving. He is also perfect meaning that there is nothing about Him this is contradictory. But How can He forgive you and me but still uphold Justice.
Look at it in this way: Judges who punish the innocent and release the wicked and quilty, are wicked, evil and are corrupt. It does not matter how much a quilty man is sorry for his crime, he must go to jail. Let's say a man raped and murdered a little girl, but is so sorry for the rape; the judge must still send him to jail. And likewise God must still punish our sins, but He can still forgive us. How? How can God forgive us and still satisfy His Justice?
This is how:
All have sinned and fall short of the glory of God. There is none righteous, no not one” (Romans 3:10, 23 NKJV).
-It means that each and every single person sins and has sinned.
“The wages of sin is death, but the gift of God is eternal life through Jesus Christ our Lord” (Romans 6:23).
-Death is eternal separation from God in hell.
- But believing in Christ alone allows you to be with God in everlasting joy.
•"The iniquity and sins of each and every persons were placed upon Jesus Christ who was sinless.
•Jesus Christ, the Son of God was crucified for our sins.
•Jesus Christ was buried and our sins were buried with Him to live no more.
•Jesus Christ arose from the dead on the third day and all who believe and receive Him as Lord arose to new life with Him.
• New life in Jesus Christ is received through faith in His death burial and resurrection, when the person repents of their sin and asks Jesus Christ to forgive their sin and wash their heart.
Beloved seek Him, read the Bible, not the catholic Bible(deutero-canonical books)
Start by read the whole of the new testament, starting from Matthew to Revelation..
i love you because Christ loves me.
@@average_loser4254 Beloved, Please give this your special attention.
God is a Good, Holy and Just( meaning that He up holds justice and does what He deems as good) Judge. He is also forgiving and He is All-Loving. He is also perfect meaning that there is nothing about Him this is contradictory. But How can He forgive you and me but still uphold Justice.
Look at it in this way: Judges who punish the innocent and release the wicked and quilty, are wicked, evil and are corrupt. It does not matter how much a quilty man is sorry for his crime, he must go to jail. Let's say a man raped and murdered a little girl, but is so sorry for the rape; the judge must still send him to jail. And likewise God must still punish our sins, but He can still forgive us. How? How can God forgive us and still satisfy His Justice?
This is how:
All have sinned and fall short of the glory of God. There is none righteous, no not one” (Romans 3:10, 23 NKJV).
-It means that each and every single person sins and has sinned.
“The wages of sin is death, but the gift of God is eternal life through Jesus Christ our Lord” (Romans 6:23).
-Death is eternal separation from God in hell.
- But believing in Christ alone allows you to be with God in everlasting joy.
•"The iniquity and sins of each and every persons were placed upon Jesus Christ who was sinless.
•Jesus Christ, the Son of God was crucified for our sins.
•Jesus Christ was buried and our sins were buried with Him to live no more.
•Jesus Christ arose from the dead on the third day and all who believe and receive Him as Lord arose to new life with Him.
• New life in Jesus Christ is received through faith in His death burial and resurrection, when the person repents of their sin and asks Jesus Christ to forgive their sin and wash their heart.
Beloved seek Him, read the Bible, not the catholic Bible(deutero-canonical books)
Start by read the whole of the new testament, starting from Matthew to Revelation..
i love you because Christ loves me.
每日一题，快快乐乐~
Nice
Good olympiad , i love the hard problem!!!
Great! Got it. :)
I usually enjoy the way you approach Math problems. However your approach here was tedious, in my humble opinion. I factorized 2009 as 7**2, 41. So, √a, √b are of the form p√41, q√41, where p+q = 7. Now read off all eight pairs.
How do you get the last step? If sqrt(a)+sqrt(b)=7sqrt(41), how do you know that sqrt(a)=p sqrt(41)?
@@TM-ht8jv Yes, you are right. Thank you for pointing this out. In order to help my argument, I must use sqrt(a) + sqrt(b) = m sqrt(r) where r is a prime number and m is an integer. Now dividing both sides by sqrt(r) and squaring we get: m**2 = (a+b)/r + 2 sqrt(ab)/r. Now I think we are in similar territory as the solution supplied by @letsthinkcritically.
انا أدعو الجميع الوقوف أمام هذه المساله طويلا
ليس طريقه حلها ولكن عمق النتائج
طبعا
nice bro
( a,b)=(7,1792)
Why have we ignored the possibility that the square roots can yield negative results too? In that case, the solution is (a,b) = (41*m^2, 41*n^2) where m+n=7 for all m,n in Integer space.
I got it at 1st try,hurray
Very well
clever man!
Certs?
Surds
Lol
Sometimes I wonder how do humans get from 1+1=2 to this
Percebeste? Percebi... Não entendi mas foi!
Can't understand this solve
Pls use more dark colour
a=0 , b=2009
😱😱 how am I so dumb
We are not dumb, they just use stupid questions
Basically, one has to know 2009 = 41*49.
Not really
You can find out that 2009= 7*2 x 41
😂😂😂
2009 = 2025 - 16
if one doesn't figure this out, gg. lol
wow
Speak properly and dont do errors
Nul
No not obvous at all🥲