here is my method of approaching 5^k-2^y=1. (actually the method used is quite similar to the method used in the video, so try it yourself!) 5^k-2^y=1 if y>=3, then using mod 8, 5^k==1 (mod 8) but since 5^2=25==1 (mod 8), then k must be even, or k=2z for positive integer z. hence, 25^z-2^y=1. 2^y=25^z-1=(5^z-1)(5^z+1) hence, 5^z-1=2^a, 5^z+1=2^b for positive integer a,b. subtracting, 2^b-2^a=2, or 2^(b-1)=2^(a-1)+1, hence a=1,b=2. hence 5^z=2^1+1=3, a contradiction, so we check cases y=1 and y=2. when y=1, 3^x=5, a contradiction. when y=2, k=1, z=2, x=2. hence (x,y,z)=(2,2,2).
I also encountered this equation a couple years ago... so I sent it to my friend who is much better at math than me and had him solve it and explain the solution to me, lmao
A suggestion for a problem (from the Dutch math olympiad) Let a, b and c be the sides of a right angled triangle, such that: a=p^m b=q^n c=2k+1 with p, q primes, and m,n and k positive integers. Find all such combinations a,b and c (Only possibilities are a=3, b=4, c=5 and a=4, b=3, c=5)
At 9:31 instead of using "magic" you could've factored 2^(y+1)=81^l-1=(3^l+1)(3^l-1)(9^l+1). As each of these factors must be a power of 2, obviously you can only have l=1, but then the last one becomes 10 so no solutions here.
mod 3 is a rather obvious guess as the 3^x term would vanish. Also, an important side effect is that 4 just happens to get reduced to 1, which has a very nice property when used as a basis in exponentiation (i.e. it just stays one). In general, when your problem involves integers, factorizations and basic number theory can help a lot to reduce the solution space. The intuition for treating stuff mod 16 comes from simple trial and error. You notice that 3^x is the dominant term in 1 + 2^(y+1) = 3^x and realize that, early on, there will be no solution if y is big enough. So we try y=1, y=2, y=3, ..., and see it won‘t work for y >= 3. Choosing y
For the gcd step can you do this? if a>b then gcd(a,b)=gcd(a-b;b) (ive seen this somewhere dunno the name. Is it Eucl. Algorithm too?) gcd(5^k + 2^y; 5^k - 2^y) -> gcd(2^(y+1); 5^k - 2^y) we can see that for y,k in natural numbers, 2^(y+1) is always even and 5^k - 2^y is always odd so gcd(~;~) = 1.
yes that's also the Euclidean algorithm (the Eucl. algorithm states that gcd(a, b) = gcd(a-kb, b) for any integers a,b,k, which follows from repeated applications of gcd(a, b) = gcd(a-b, b). If you know what modular arithmetic is then the Eucl. algorithm can also be stated as gcd(a, b) = gcd(a mod b, b).)
If you plug in a specific y into 1 + 2^(y+1) = 3^x, your equation essentially becomes 3^x = c, for some positive constant c. Assuming x to be a real number for now, there‘s only one solution to this as the exponential function is stricly increasing (analysis/calculus 1). As this is true for all real numbers x, it is also true for all x in IN or Z.
We can make something much more difficult: Find all natural number a,b,c such that: 2^a=3^b+5^c In my opinion this problem can be easily a problem at IMO this years
I applied the Am-Gm. It goes like, as LHS &rhs >0 then You see for x=4n and y=2n+1 the addition of LHS numbers contains last digit =5 so we can compare it with rhs ( it has only last digit =5 :) Now,go for am-gm you get 25^z> 12.48²ⁿ as you put this x,y I made inside this. Then go for z=0,1,2,.... You get when z>3 no solution actually exists :) then (X,Y,Z)=(0,1,1) & (2,2,2) that's it.👍
There always seem to be two interpretations of Natural numbers, and I prefer including 0 for two reasons. A) Natural numbers arise from the context of counting, and zero can naturally appear in that context. E.g. as an introductory exercise in statistics (tallying, frequency tables, mode, median, mean) I would ask each of a class of young students how many brothers and sisters they had in their family. For some, the answer would be zero. B) We have the Z+ notation for the positive integers, so it seems sensible to use the N notation for a different purpose.
Its more of a general case if this problem. Instead of the bases being a particular Pythagorean triplet, the result has to be proved for any arbitrary a,b,c belonging to N.
For 2^(y+1)=3^x - 1: You could instead notice that y=1 is not a solution so y + 1 is at least 3 and the LHS is congruent to 0 (mod 8) hence also the RHS. If x were odd then 3^x - 1 == 2 (mod 8) which is a contradiction so x is even. Let x=2k. Now: 2^(y+1)=(3^k + 1)(3^k - 1) One of 3^k±1 is congruent to 2 (mod4) but the only power of two congruent to 2(mod4) is 2. So 3^k - 1=2, 3^k+1=4 so k=1 and x=2 and y=2 and hence z=2
Great! Very similar to my solution. I said, if x is odd, 3^x - 1 ≡ 2 (mod 4), so 2^(y+1) ≡ 2 (mod 4), therefore y+1=1 & y=0 (disallowed value). If x is even, say x=2k, therefore 2^(y+1) = (3^k-1)(3^k+1), so both those factors of a power of 2 are themselves powers of 2. But the difference between them is 2, so they are 2 and 4, and the conclusion follows.
This question was released before Fermat's Last Theorem was proven. It is probably unlikely now that a similar problem would be posed in a competition scenario. Would it suffice though to cite the modularity theorem if one is asked a similar question?
Let n be a positive integer. Show that any number greater than n⁴/16 can be written in at most one way as the product of two of its divisors having difference not exceeding n. (1998 St. Petersburg City Mathematical Olympiad) Please solve this one. Thanks.
a3=2, a2, a1, a0. There are 6 combinations of 2 positions in the 4 digit number where exactly 2 digits can be equal. Because 4C2 = 3! = 6 Of those 6 combinations 2 cases arise. 1st: One position is shared by the fixed first digit(i.e. 2) and second vice versa. For 1st case: There is 3 such possibilities. a3=am=2. 0≤m≤2 So, for each m there is n such that 0≤n≤2 and n≠m. There are 2 such n, lets call them n1,n2. There is only 9×8 permutation for such a(n1),a(n2). Because, a(n1)≠a(n2)≠2. Since, there are 3 possible values of m. Total permutation is 3*9*8. Case 2: There is 6-3=3 possibilities in this case. a(m1)=a(m2)≠2. Where 0≤m1,m2≤2. m1≠m2. There are 9 possibilities. For the one remaining n. a(n)≠a(m1)≠2. Hence, 8 such possibilities. Total permutations=3*9*8. Summing up total permutations=3*9*8+3*9*8 =432 Is the answer
dude from fermat's last theorem we may conclude that x0 ,then one more possible solution is (x,y,z)=(0,1,1) ,which can't be accepted in above question but it is valid.
Fermat last theorem is about the equation a^n + b^n = c^n, ie when all exponents are the same It says nothing about a more general equation like this one
I must confess that this type of problem and the algebraic solution given here has no appeal, and I cannot see its being in any way elegant. [ Clever yes, and laborious at the same time ]. Of much greater appeal might be if a solution based on a geometric analogue could be found. The similarity to Fermat's last theorem is obvious, and accordingly that alone would inhibit me from attempting this problem. Being myself untutored I have no hope of ever following Weil's proof of Fermat. To judge by the disparity between Weil's complexity and the deceptive simplicity of the Fermat proposition, in the back of my mind is the suspicion that the solution given here might contain some flaw.
this question is not similar to Fermat's last theorem - in Fermat's last theorem the three exponents are equal and the three bases are arbitrary (whereas here the three exponents can be different and the three bases are given)
here is my method of approaching 5^k-2^y=1.
(actually the method used is quite similar to the method used in the video, so try it yourself!)
5^k-2^y=1
if y>=3, then using mod 8,
5^k==1 (mod 8)
but since 5^2=25==1 (mod 8),
then k must be even, or k=2z for positive integer z.
hence, 25^z-2^y=1.
2^y=25^z-1=(5^z-1)(5^z+1)
hence, 5^z-1=2^a, 5^z+1=2^b for positive integer a,b.
subtracting, 2^b-2^a=2, or 2^(b-1)=2^(a-1)+1, hence a=1,b=2.
hence 5^z=2^1+1=3, a contradiction, so we check cases y=1 and y=2.
when y=1, 3^x=5, a contradiction.
when y=2, k=1, z=2, x=2. hence (x,y,z)=(2,2,2).
I remember solving this equation a couple years ago... i was so hyped at the time from my solution
I also encountered this equation a couple years ago... so I sent it to my friend who is much better at math than me and had him solve it and explain the solution to me, lmao
@@replicaacliper hahahaha
A suggestion for a problem (from the Dutch math olympiad)
Let a, b and c be the sides of a right angled triangle, such that:
a=p^m
b=q^n
c=2k+1
with p, q primes, and m,n and k positive integers. Find all such combinations a,b and c
(Only possibilities are a=3, b=4, c=5 and a=4, b=3, c=5)
At 9:31 instead of using "magic" you could've factored 2^(y+1)=81^l-1=(3^l+1)(3^l-1)(9^l+1). As each of these factors must be a power of 2, obviously you can only have l=1, but then the last one becomes 10 so no solutions here.
May i ask how you got the intuition to use modulos such as mod 16, mod 5 and mod 3?
mod 3 is a rather obvious guess as the 3^x term would vanish. Also, an important side effect is that 4 just happens to get reduced to 1, which has a very nice property when used as a basis in exponentiation (i.e. it just stays one). In general, when your problem involves integers, factorizations and basic number theory can help a lot to reduce the solution space.
The intuition for treating stuff mod 16 comes from simple trial and error. You notice that 3^x is the dominant term in 1 + 2^(y+1) = 3^x and realize that, early on, there will be no solution if y is big enough. So we try y=1, y=2, y=3, ..., and see it won‘t work for y >= 3. Choosing y
For the gcd step can you do this?
if a>b then gcd(a,b)=gcd(a-b;b) (ive seen this somewhere dunno the name. Is it Eucl. Algorithm too?)
gcd(5^k + 2^y; 5^k - 2^y) -> gcd(2^(y+1); 5^k - 2^y) we can see that for y,k in natural numbers, 2^(y+1) is always even and 5^k - 2^y is always odd so gcd(~;~) = 1.
yes that's also the Euclidean algorithm (the Eucl. algorithm states that gcd(a, b) = gcd(a-kb, b) for any integers a,b,k, which follows from repeated applications of gcd(a, b) = gcd(a-b, b). If you know what modular arithmetic is then the Eucl. algorithm can also be stated as gcd(a, b) = gcd(a mod b, b).)
Really nice video
Excellent work, as usual
im a little confused. how do you know there's only one solution for a specific y value? did I miss something?
If you plug in a specific y into 1 + 2^(y+1) = 3^x, your equation essentially becomes 3^x = c, for some positive constant c. Assuming x to be a real number for now, there‘s only one solution to this as the exponential function is stricly increasing (analysis/calculus 1). As this is true for all real numbers x, it is also true for all x in IN or Z.
We can make something much more difficult:
Find all natural number a,b,c such that:
2^a=3^b+5^c
In my opinion this problem can be easily a problem at IMO this years
a similar question could be 5^a+12^b=13^c
Amazing solution. 👍👍👍
I applied the Am-Gm.
It goes like, as LHS &rhs >0 then
You see for x=4n and y=2n+1 the addition of LHS numbers contains last digit =5 so we can compare it with rhs ( it has only last digit =5 :)
Now,go for am-gm you get 25^z> 12.48²ⁿ as you put this x,y I made inside this.
Then go for z=0,1,2,.... You get when z>3 no solution actually exists :) then
(X,Y,Z)=(0,1,1) & (2,2,2) that's it.👍
Fermat's last theorem if x=y=z
I used a similar solution!
So we are assuming x,y,z are all positive then, else (0,1,1) would be another solution.
Yes, because 0 is not a natural number.
@@letsthinkcritically Sometimes people consider 0 to be a natural number.
@@letsthinkcritically in Turkish curriculum we consider 0 a natural number.
@@letsthinkcritically don't hurt my feelings :0
There always seem to be two interpretations of Natural numbers, and I prefer including 0 for two reasons.
A) Natural numbers arise from the context of counting, and zero can naturally appear in that context. E.g. as an introductory exercise in statistics (tallying, frequency tables, mode, median, mean) I would ask each of a class of young students how many brothers and sisters they had in their family. For some, the answer would be zero.
B) We have the Z+ notation for the positive integers, so it seems sensible to use the N notation for a different purpose.
Hi @letsthinkcritically isn't this somewhat related to fermats last theorem
Its more of a general case if this problem. Instead of the bases being a particular Pythagorean triplet, the result has to be proved for any arbitrary a,b,c belonging to N.
You’re gcd notation is confusing, I’d recommend writing gcd outside th brackets if two no.s. Just advice tho, great vid!!!
I solved that question 5 hours before upload. I didn't know it was IMO shortlist. I thought it was too easy.
Nice work. but, What about solution x=0; y=1; z=1 ?
x,y,z are natural numbers
nice problem :D
For 2^(y+1)=3^x - 1:
You could instead notice that y=1 is not a solution so y + 1 is at least 3 and the LHS is congruent to 0 (mod 8) hence also the RHS.
If x were odd then 3^x - 1 == 2 (mod 8) which is a contradiction so x is even. Let x=2k. Now:
2^(y+1)=(3^k + 1)(3^k - 1)
One of 3^k±1 is congruent to 2 (mod4) but the only power of two congruent to 2(mod4) is 2. So
3^k - 1=2, 3^k+1=4 so k=1 and x=2 and y=2 and hence z=2
Great! Very similar to my solution. I said, if x is odd, 3^x - 1 ≡ 2 (mod 4), so 2^(y+1) ≡ 2 (mod 4), therefore y+1=1 & y=0 (disallowed value).
If x is even, say x=2k, therefore 2^(y+1) = (3^k-1)(3^k+1), so both those factors of a power of 2 are themselves powers of 2. But the difference between them is 2, so they are 2 and 4, and the conclusion follows.
This question was released before Fermat's Last Theorem was proven. It is probably unlikely now that a similar problem would be posed in a competition scenario. Would it suffice though to cite the modularity theorem if one is asked a similar question?
Fermat's Last Theorem is only relevant to the case x=y=z here so I wouldn't say this question is similar to FLT
(x,y,z)=(2,2,2) isn’t the only solution to the problem. (0,1,1) also works.
Nice job, though!
@@jimjim3979 See the article on *natural number* in the Wikipedia.
0 is not natural number
Let n be a positive integer. Show that any number greater than n⁴/16 can be written in at most one way as the product of two of its divisors having difference not exceeding n.
(1998 St. Petersburg City Mathematical Olympiad)
Please solve this one.
Thanks.
All ik is that fermats last theorem states that x^a+y^a=z^a
is true if a is 2
I think that it's very easy but it is 1991.
Did you just prove Fermat's Last Theorem?
Only for the 3,4 and 5 case
@@caiocesar2093 I see.
try this
how many numbers are there between 2000 and 2999 such that exactly two of its digits are same?
a3=2, a2, a1, a0.
There are 6 combinations of 2 positions in the 4 digit number where exactly 2 digits can be equal.
Because 4C2 = 3! = 6
Of those 6 combinations 2 cases arise.
1st: One position is shared by the fixed first digit(i.e. 2) and second vice versa.
For 1st case: There is 3 such possibilities.
a3=am=2. 0≤m≤2
So, for each m there is n such that 0≤n≤2 and n≠m.
There are 2 such n, lets call them n1,n2.
There is only 9×8 permutation for such a(n1),a(n2).
Because, a(n1)≠a(n2)≠2.
Since, there are 3 possible values of m. Total permutation is 3*9*8.
Case 2:
There is 6-3=3 possibilities in this case.
a(m1)=a(m2)≠2. Where 0≤m1,m2≤2.
m1≠m2.
There are 9 possibilities.
For the one remaining n. a(n)≠a(m1)≠2.
Hence, 8 such possibilities.
Total permutations=3*9*8.
Summing up total permutations=3*9*8+3*9*8
=432
Is the answer
stie iustin cealalta metoda
dude from fermat's last theorem we may conclude that x0 ,then one more possible solution is
(x,y,z)=(0,1,1) ,which can't be accepted in above question but it is valid.
Fermat last theorem is about the equation a^n + b^n = c^n, ie when all exponents are the same
It says nothing about a more general equation like this one
This is for Beal's conjecture, not Fermat's Last Theorem
WRONG, you provided only half of the answer (you missed by 50%)! (0, 1, 1) is the other solution since 3^0 + 4^1 = 5^1
x, y and z are natural.
I must confess that this type of problem and the algebraic solution given here has no appeal, and I cannot see its being in any way elegant. [ Clever yes, and laborious at the same time ]. Of much greater appeal might be if a solution based on a geometric analogue could be found.
The similarity to Fermat's last theorem is obvious, and accordingly that alone would inhibit me from attempting this problem.
Being myself untutored I have no hope of ever following Weil's proof of Fermat. To judge by the disparity between Weil's complexity and the deceptive simplicity of the Fermat proposition, in the back of my mind is the suspicion that the solution given here might contain some flaw.
this question is not similar to Fermat's last theorem - in Fermat's last theorem the three exponents are equal and the three bases are arbitrary (whereas here the three exponents can be different and the three bases are given)