# 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
2022 Award
You mean a proof for ##5-1=2^2?##

• WWGD, mfb and dextercioby
Homework Helper
Gold Member
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$$10^3 - 6^3 = 784 = 28^2$$

• WWGD, mfb and dextercioby
Homework Helper
Gold Member
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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
Gold Member
2022 Award
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
2022 Award
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
Homework Helper
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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 $$c$$