# Proof of Differences of Odd Powers

• I
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?

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

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

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

dextercioby
e2m2a
$$6^5 - 2^5 = 88^2$$
Ha Ha! You are good! How do you find these solutions so quickly?

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

Gold Member
$$6^5 - 2^5 = 88^2$$
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?

Mentor
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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
Gold Member
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:
Mentor
##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
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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
$$8^3-7^3=512-343=169=13^2$$
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 $$c$$