# Proof of Differences of Odd Powers

• I
• e2m2a
In summary, the conversation discusses finding proofs for the impossibility of an odd powered integer difference being equal to a square. Examples are given, and it is mentioned that Wiles has proven this. The discussion also touches on the relationship between integer solutions and the proof by Wiles. Finally, it is shown that there exists a solution for certain values of n and a and b.

#### 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?

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

WWGD, mfb and dextercioby
$$10^3 - 6^3 = 784 = 28^2$$

WWGD, mfb and dextercioby
$$6^5 - 2^5 = 88^2$$

dextercioby
PeroK said:
$$6^5 - 2^5 = 88^2$$
Ha Ha! You are good! How do you find these solutions so quickly?

e2m2a said:
Ha Ha! You are good! How do you find these solutions so quickly?
Just used a spreadsheet.

PeroK said:
$$6^5 - 2^5 = 88^2$$
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?

bob012345 said:
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.

WWGD, dextercioby and bob012345
fresh_42 said:
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##?

Last edited:
##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.

dextercioby
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##.

Last edited:
bob012345 and Office_Shredder
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 $$c$$