I did it by considering that xy - 7 > x + y > sqrt(x^2+y^2) for all x,y >= 4. Therefore, if a solution exists, at least one of x,y must be less than 4. Direct calculation of the remaining cases revealed 0,7 7,0 3,4 4,3.
If we write this in a quadratic equations in X, the discriminant must be a perfect square and the result will be much easy as P^2-Y^2=48. After that is very easy. Divisors of 48...
I know this could be solved algebraically, as he showed us, but I thought of differential equations to help solve this. His second solution used an introduction of two variables "a" and "b" to solve for "x" and "y" but I would have introduced implicit differentiation and differential equations from calculus. The squares would reduce from second degree to first degree terms, while instead of variables a,b I would have dy/dx ( or y') and dx/dy (or x') and the constant 7 would disappear, since derivatives are essentially about finding slopes of curves. But, I'm sure that maybe 🤔 only algebra should have been used for this to enhance algebraically thinking skills and advanced logical thinking on algebra and geometry terms.
For all integers x and y bigger than 1 xy is bigger than x+y .in this question x²+y²=(xy-7)² and we know that x²+y² is less than or equal to (x+y)² then you take sqrt of both xy-7 squared and x+y squared and for all cases that xy is bigger than 7 we get (xy-7)
I believe the Pythagorean part is wrong. I believe he is saying that since (x, y, xy - 7) is a Pythagorean triple, there exists a pair of integers (a, b) such that (x, y, xy - 7) = (a^2 - b^2, 2ab, a^2 + b^2). Euclid’s formula for generating Pythagorean triples states that (a^2 - b^2, 2ab, a^2 + b^2) is always a Pythagorean triple but does not generate all of them (Wikipedia says that (9, 12, 15) can’t be generated). I believe he is able to get the answer this way because the answers happened to be cases that the Euclid’s formula does not miss.
Another solution. Note if x,y are even or x,y are odds Mod 4 doesn't work. Then we can assume that x is even and y is odd. Let r= x*y. If r=6 or 8 then x=0 and x^2=1, That doesn't work. Note that (x+y)^2=x^2+y^2+2*x*y=(r-7)^2+2*r. And (x-y)^2=x^2+y^2-2*x*y=(r-7)^2-2*r. Thus (r-7)^2+-2*r are square. If r>8 then (r-8)^2=(r-7+1)^2=(r-7)^2+2*(r-7)+1=(r-7+2)=(r-7)^2+4*(r-7)+4=(r-7)^2+4*r-28+4. Thus 24>=2*r. r=12)). Cases 0
I initially tried the Pythagorean triples approach but then hit a stumbling block that you glossed over. The quoted formula giving (x,y,xy-7) in terms of (a,b) only generates primitive triples, which (x,y,xy-7) is not guaranteed to be (because they can share a factor of 7, for example). So your solution here is a bit incomplete unfortunately. To deal with the remaining case where say x=7k, y=7l, we're led to consider the equation k^2+l^2=(7kl-1)^2. I gave up at this point because this was starting to get a bit too long. Is there a quick way to salvage the Pythagorean triples approach?
" Is there a quick way to salvage the Pythagorean triples approach? " For a grubby and unrepentant pragmatist like me - YES. By inspection, start with either x =3, y = 4 or vice versa. then x*y = 12 and 12-7 = 5, giving the most common and best known PT. Oh and BTW, I personally prefer geometric solutions to apparently algebraic problems, for the simple reason that they show a real-world meaning as opposed to the purely abstract.
11:00 i didn't understand a single thing lol but I think that assuming is kinda risky , maybe this time not but other times it's not the most accurate thing
Step 1. We know if given a right triangle then a^2 +b^2 = c^2. That's how the Pythagorean theorem is typically used. But it's also worth noting that If a^2 +b^2 = c^2 then there exists a triangle with a right angle such that C^2 is the hypotenuse. It's the converse of the Pythagorean theorem and also in Euclid's Elements. Step 2. We know that because such a triangle exists, and we're using integers only for a, b, c then those integers must be a Pythagorean triple. It's how a Pythagorean Triple is defined. And by Euclid's formula for generating Pythagorean triples (editing to add: googling that formula will explain far better than I can) we know there are these other integers m and n where m>n>0, a=m^2-n^2, b=2mn, and c=m^2+n^2. Step 3 is what is shown in the video, the combining of those established definitions to get to the set of equations he uses to solve for the 3,4 portion. Not exactly what I did but it works. The 0,7 pair is so immediately obvious that it makes the 3,4 look a lot more complicated but it's not, it's only when adding rigor to the solution that things get trickier than they need to be, imho. Hope that helped!
Harry Sakata 0 seconds ago I believe the Pythagorean part is wrong. I believe he is saying that since (x, y, xy - 7) is a Pythagorean triple, there exists a pair of integers (a, b) such that (x, y, xy - 7) = (a^2 - b^2, 2ab, a^2 + b^2). Euclid’s formula for generating Pythagorean triples states that (a^2 - b^2, 2ab, a^2 + b^2) is always a Pythagorean triple but does not generate all of them (Wikipedia says that (9, 12, 15) can’t be generated). I believe he is able to get the answer this way because the answers happened to be cases that the Euclid’s formula does not miss.
@@thane9 Euclid's formula does not claim that all Pythagorean triples can be expressed using Euclid's formula. I think he arrives at the answer this way because the answers just happened to be ones that Euclid does not miss.
Hello, I thank you for your beautifully performed processing, solution, and video. Fifty-one years ago, I used to do the math and did integrals for my girlfriend to solve and find the answers to become like, "I love you, I am yours, the life is beautiful, you are gorgeous, ETC. For reasons of the universe's laws of physic, mathematics or mathematicians always mimic physic to achieve formulas or formats doing what physic and or logic does and is accurate. The equation in this video made me curious to solve it my way, and it gave me more answers than expected; surprisingly, the equation has many solutions: y=6.0e⁰, x=1.66282713421e-¹ x=6.0e⁰, y=2.23371728658e⁰ y=1.1e¹, x=1.64751790548e⁰ x=1.1e¹, y=1.64751790548e⁰ y=2.8e¹, x=1.25099758593e⁰ x=2.8e¹, y=1.25099758593e⁰ y=1.090e², x=1.642678947e⁰ x=1.090e². y=1.642678947e⁰ y=3.0e, x=1.0233338915e⁰ x=3.0e³, y=1.0233338915e⁰ y=7.0e⁰, x=0.0e⁰ x=7, y=2.04166666667e⁰ y=1.0e⁵, x=1.00007000005e⁰ x=1.0e⁵, y=1.00007000005e⁰ y=5.0e¹, x=4.6879240832e⁰ x=5.0e¹, y=4.6879240832e⁰ y=1.36893e⁰, x=2.82216581448e⁰ x=2.82216581448e⁰, y=2.82216581448e⁰ y=0.0e⁰, x=7.0e⁰ x=0.0e⁰, y=7.0e⁰ y=2.063250108030.0e⁰, x=2.0e⁰ x=2.0e⁰, y=7.270083225280.0e⁰ y=3.0e⁰, x=2.0632501803.0e⁰ x=3.00.0e⁰, y=1.250.0e⁰ y=1.0e⁶, x=1.00000700000e⁶ x=1.0e⁶, y=1.00000700000e⁶ y=4.0e⁰, x=7.33333333333e-¹ x=4.0e⁰, y=2.999999999999e⁰ y=1.00000000000e-8, x=6.99999993000e⁶ x=1.00000000000e-8, y=6.99999993000e⁶ y=7.00000000000e-4, x=6.99510669260e⁰ x=7.00000000000e-4, y=6.99510339260e⁰ y=1.01897400249e⁰, x=3.39295773262e⁰ x=3.69000000000e², y=1.01897400249e⁰ ETC. For instance, the value of angles of the triabgle with x=3.69e², y=1.01897400249e⁰, and (xy-7.0e¹=3.69001406919e²) sides, are 9.0e¹°, 0.1582188808888e-¹°, and 8.98417811358e¹°; 9.0e¹°+0.1582188808888e-¹°+8.98417811358e¹°=1.80e²° Furthermore, I am neither trying to be an impertinent nor a rude person. Very best regards,
There are infinitely many pairs of x,y that satisfy the equation - but there are only four when you specify x,y to be non-negative integers, as in the problem.
I did it by considering that xy - 7 > x + y > sqrt(x^2+y^2) for all x,y >= 4. Therefore, if a solution exists, at least one of x,y must be less than 4. Direct calculation of the remaining cases revealed 0,7 7,0 3,4 4,3.
What a channel this is!! I absolutely love your content and wish I could be as good as you in this algebraic gymnastics ☺️😊
The Pythagorean triple part was very well thought out.
If we write this in a quadratic equations in X, the discriminant must be a perfect square and the result will be much easy as P^2-Y^2=48. After that is very easy. Divisors of 48...
This inspired my own similar problem. What are the integer solutions to x^4 - 9x^2 - 2xy - y^2 = 73 ?
I know this could be solved algebraically, as he showed us, but I thought of differential equations to help solve this. His second solution used an introduction of two variables "a" and "b" to solve for "x" and "y" but I would have introduced implicit differentiation and differential equations from calculus. The squares would reduce from second degree to first degree terms, while instead of variables a,b I would have dy/dx ( or y') and dx/dy (or x') and the constant 7 would disappear, since derivatives are essentially about finding slopes of curves. But, I'm sure that maybe 🤔 only algebra should have been used for this to enhance algebraically thinking skills and advanced logical thinking on algebra and geometry terms.
We can also use the inequality :(xy-7)
How can you be so sure that x+y>=xy-7?
@@niloneto1608 that inequality is wrong
@@motherisape no it’s not. They just flipped the sides
For all integers x and y bigger than 1 xy is bigger than x+y .in this question x²+y²=(xy-7)² and we know that x²+y² is less than or equal to (x+y)² then you take sqrt of both xy-7 squared and x+y squared and for all cases that xy is bigger than 7 we get (xy-7)
Please think before you judge... I proved it in the comments replied to @Nilo Neto
I initially thought ellipse and reactangle hyperbola formula
(xy -7 -x)(xy -7 + x) = y^2
case I :
xy -7 -x = 1
xy -7 + x = y^2
ie 2x = (y-1)(y+1)
y -1 = 2 , y +1 = x
y = 3 , x = 4
Note that (x,y) is solution if only if (y ,x) is solution, if only if (-x,-y) is solution. Then we can assume x>=y and x>0.
0:44 why do we add +2xy ?
To make x2 +y2 as a perfect square as (x+y)²
I used the first approach. After some initial dead ends with differences of other squares. The Pythagorean approach is interesting.
Can you explain the Pythagoras part?
I believe the Pythagorean part is wrong.
I believe he is saying that since (x, y, xy - 7) is a Pythagorean triple, there exists a pair of integers (a, b) such that (x, y, xy - 7) = (a^2 - b^2, 2ab, a^2 + b^2). Euclid’s formula for generating Pythagorean triples states that (a^2 - b^2, 2ab, a^2 + b^2) is always a Pythagorean triple but does not generate all of them (Wikipedia says that (9, 12, 15) can’t be generated). I believe he is able to get the answer this way because the answers happened to be cases that the Euclid’s formula does not miss.
asnwer=4 my munber isit
Why is the graph so cute? (Plot it!)
You should problem 4 from the same year, very fun!
At the beginning of the video I think you said for negative integers x & y
"non-negative" This is what people say when they want to allow 0 also.
Great! I got it. :)
i have solved it in a better way
Another solution. Note if x,y are even or x,y are odds Mod 4 doesn't work. Then we can assume that x is even and y is odd. Let r= x*y. If r=6 or 8 then x=0 and x^2=1, That doesn't work. Note that (x+y)^2=x^2+y^2+2*x*y=(r-7)^2+2*r. And (x-y)^2=x^2+y^2-2*x*y=(r-7)^2-2*r.
Thus (r-7)^2+-2*r are square. If r>8 then (r-8)^2=(r-7+1)^2=(r-7)^2+2*(r-7)+1=(r-7+2)=(r-7)^2+4*(r-7)+4=(r-7)^2+4*r-28+4. Thus 24>=2*r. r=12)). Cases 0
I initially tried the Pythagorean triples approach but then hit a stumbling block that you glossed over. The quoted formula giving (x,y,xy-7) in terms of (a,b) only generates primitive triples, which (x,y,xy-7) is not guaranteed to be (because they can share a factor of 7, for example). So your solution here is a bit incomplete unfortunately. To deal with the remaining case where say x=7k, y=7l, we're led to consider the equation k^2+l^2=(7kl-1)^2. I gave up at this point because this was starting to get a bit too long.
Is there a quick way to salvage the Pythagorean triples approach?
" Is there a quick way to salvage the Pythagorean triples approach? "
For a grubby and unrepentant pragmatist like me - YES. By inspection, start with either x =3, y = 4 or vice versa. then x*y = 12 and 12-7 = 5, giving the most common and best known PT.
Oh and BTW, I personally prefer geometric solutions to apparently algebraic problems, for the simple reason that they show a real-world meaning as opposed to the purely abstract.
Using Pythagorean triplet a²+b²=c², with a=3 and b=4 then c=5. If x=3 and y=4, xy=12 --> xy-7=5, giving x²+y²=(xy-7)². Thus (x,y)=(3,4)
Good one, maybe slightly harder problems are better
11:00 i didn't understand a single thing lol but I think that assuming is kinda risky , maybe this time not but other times it's not the most accurate thing
Yeah, I didn't understand the 2nd approach, but the first one does make sense
That was really good.
Take x=0 , y=7
x=7 y = 0
Could you explain the Pythagoras part? Which knowleadge does it use? Why exists a,b as integers since that is a right triangle?
Step 1. We know if given a right triangle then a^2 +b^2 = c^2. That's how the Pythagorean theorem is typically used. But it's also worth noting that If a^2 +b^2 = c^2 then there exists a triangle with a right angle such that C^2 is the hypotenuse. It's the converse of the Pythagorean theorem and also in Euclid's Elements.
Step 2. We know that because such a triangle exists, and we're using integers only for a, b, c then those integers must be a Pythagorean triple. It's how a Pythagorean Triple is defined. And by Euclid's formula for generating Pythagorean triples (editing to add: googling that formula will explain far better than I can) we know there are these other integers m and n where m>n>0, a=m^2-n^2, b=2mn, and c=m^2+n^2.
Step 3 is what is shown in the video, the combining of those established definitions to get to the set of equations he uses to solve for the 3,4 portion.
Not exactly what I did but it works. The 0,7 pair is so immediately obvious that it makes the 3,4 look a lot more complicated but it's not, it's only when adding rigor to the solution that things get trickier than they need to be, imho. Hope that helped!
Harry Sakata
0 seconds ago
I believe the Pythagorean part is wrong.
I believe he is saying that since (x, y, xy - 7) is a Pythagorean triple, there exists a pair of integers (a, b) such that (x, y, xy - 7) = (a^2 - b^2, 2ab, a^2 + b^2). Euclid’s formula for generating Pythagorean triples states that (a^2 - b^2, 2ab, a^2 + b^2) is always a Pythagorean triple but does not generate all of them (Wikipedia says that (9, 12, 15) can’t be generated). I believe he is able to get the answer this way because the answers happened to be cases that the Euclid’s formula does not miss.
@@thane9 Euclid's formula does not claim that all Pythagorean triples can be expressed using Euclid's formula. I think he arrives at the answer this way because the answers just happened to be ones that Euclid does not miss.
2+6=?
8
26
2+2 = 5
ok
Mate U lost 2xy in the right Side ???
You don't know algebra
x^2+2xy+y^2 factors to be (x+y)^2. That’s were 2xy went
😂😂
@@motherisape Explain to him like Dwight did instead of being an ass.
@@motherisape he's white or american so you know why 😂
Hello,
I thank you for your beautifully performed processing, solution, and video. Fifty-one years ago, I used to do the math and did integrals for my girlfriend to solve and find the answers to become like, "I love you, I am yours, the life is beautiful, you are gorgeous, ETC. For reasons of the universe's laws of physic, mathematics or mathematicians always mimic physic to achieve formulas or formats doing what physic and or logic does and is accurate.
The equation in this video made me curious to solve it my way, and it gave me more answers than expected; surprisingly, the equation has many solutions:
y=6.0e⁰, x=1.66282713421e-¹
x=6.0e⁰, y=2.23371728658e⁰
y=1.1e¹, x=1.64751790548e⁰
x=1.1e¹, y=1.64751790548e⁰
y=2.8e¹, x=1.25099758593e⁰
x=2.8e¹, y=1.25099758593e⁰
y=1.090e², x=1.642678947e⁰
x=1.090e². y=1.642678947e⁰
y=3.0e, x=1.0233338915e⁰
x=3.0e³, y=1.0233338915e⁰
y=7.0e⁰, x=0.0e⁰
x=7, y=2.04166666667e⁰
y=1.0e⁵, x=1.00007000005e⁰
x=1.0e⁵, y=1.00007000005e⁰
y=5.0e¹, x=4.6879240832e⁰
x=5.0e¹, y=4.6879240832e⁰
y=1.36893e⁰, x=2.82216581448e⁰
x=2.82216581448e⁰, y=2.82216581448e⁰
y=0.0e⁰, x=7.0e⁰
x=0.0e⁰, y=7.0e⁰
y=2.063250108030.0e⁰, x=2.0e⁰
x=2.0e⁰, y=7.270083225280.0e⁰
y=3.0e⁰, x=2.0632501803.0e⁰
x=3.00.0e⁰, y=1.250.0e⁰
y=1.0e⁶, x=1.00000700000e⁶
x=1.0e⁶, y=1.00000700000e⁶
y=4.0e⁰, x=7.33333333333e-¹
x=4.0e⁰, y=2.999999999999e⁰
y=1.00000000000e-8, x=6.99999993000e⁶
x=1.00000000000e-8, y=6.99999993000e⁶
y=7.00000000000e-4, x=6.99510669260e⁰
x=7.00000000000e-4, y=6.99510339260e⁰
y=1.01897400249e⁰, x=3.39295773262e⁰
x=3.69000000000e², y=1.01897400249e⁰
ETC.
For instance, the value of angles of the triabgle with x=3.69e², y=1.01897400249e⁰, and (xy-7.0e¹=3.69001406919e²) sides, are 9.0e¹°, 0.1582188808888e-¹°, and 8.98417811358e¹°; 9.0e¹°+0.1582188808888e-¹°+8.98417811358e¹°=1.80e²°
Furthermore, I am neither trying to be an impertinent nor a rude person.
Very best regards,
There are infinitely many pairs of x,y that satisfy the equation - but there are only four when you specify x,y to be non-negative integers, as in the problem.
@@EssentialsOfMath
Hello,
There are many more than four (4) nonnegative agents listed in the calculation.
Regards,
Algebra is love ....thus guy is doing great have a look #mathmarvelasmr