Proof of Differences of Odd Powers

[e2m2a]

TL;DR Summary

Researching for proof for differences of powers.

I am interested in finding any proofs that exist which demonstrates that the difference between two odd powered integers can never be equal to a square? Has there been any research in this? For example, given this expression a^n -b^n = c^2, where a,b,c are positive integers and a>b, n = odd power. Has there been any proof that an integer solution for this is impossible?

 

[fresh_42]

You mean a proof for $5-1=2^2?$

 

[PeroK]

$$10^3 - 6^3 = 784 = 28^2$$

 

[PeroK]

$$6^5 - 2^5 = 88^2$$

 

[e2m2a]

$$6^5 - 2^5 = 88^2$$

Ha Ha! You are good! How do you find these solutions so quickly?

 

[PeroK]

Ha Ha! You are good! How do you find these solutions so quickly?

Just used a spreadsheet.

 

[bob012345]

$$6^5 - 2^5 = 88^2$$

For every integer that is a solution of $a^n - b^n = c^2$, can you prove $c^2 ≠ d^n$ where d is another integer?

 

[fresh_42]

For every integer that is a solution of $a^n - b^n = c^2$, can you prove $c^2 ≠ d^n$ where d is another integer?

Yes. Wiles did.

 

[bob012345]

Yes. Wiles did.

I know but wait, isn't proving $d^n$ can't be some integer squared less than what Wiles did that there is no $d^n$ that satisfies $a^n - b^n$?

 

[mfb]

$a^n - b^n = d^n$ implies $a^n + d^n = b^n$ which has no non-trivial integer solutions for n>2. For n=1 solutions are trivial and for n=2 there are solutions.

 

[PeroK]

For any odd $n$, consider $b = 2^n -1$ and $a = 2(2^n-1)$. Then:
$$a^n - b^n = 2^n(2^n - 1)^n - (2^n-1)^n = (2^n - 1)^{n+1}$$This is a square as $n + 1$ is even.

Also, if $a, b, c$ is a solution, then so is $k^2a, k^2b, k^nc$.

 

[WWGD]

Sorry, just to continue the pile up:
$$ 8^3-7^3=512-343=169=13^2$$
Edit: But notice you can use $$a=a'^2, b=b'^2$$, then:
$$a^n-b^n=(a'^n)^2-(b'^n)^2$$ for which may equal $$c^2$$ for some integer $$c$$