# a theorem of Fermat

While I was surfing the web, searching things about math, I read something about a particular theorem of Fermat. It said: let $a$ and $b$ be rational. Then $a^4-b^4$ cannot equal the square of a rational number, so $a^4-b^4\neq c^2$ with $c$ rational. My question is: did I understand the theorem? if not, what is the actual theorem? if yes, how can I proof this or is that too difficult?

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Yes, this is a theorem of Fermat, in the equivalent version that there are no non-trivial integer solutions. The proof is by Fermat's Method of Infinite Descent, a version of strong induction. It is one of the few number theoretic assertions of Fermat for which he sketched enough of a proof for us to know for sure how he did it. – André Nicolas Oct 29 '12 at 15:39
See my speculative reconstruction of Fibonacci's Lost Theorem $= \rm FLT_4$. – Bill Dubuque Oct 29 '12 at 17:45

That's right, modulo a small precision: Fermat proved that there are no solutions other than the obvious ones, such as $(a,b,c)=(0,0,0)$, $(1,0,1)$, etc.. See for instance this wikipedia page.

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Yes you understand the theorem. The proof uses the method of "infinite descent" (a sibling of reductio ad absurdum, or proof by contradiction).

There is a walkthrough proof here:- https://en.wikipedia.org/wiki/Proof_of_Fermat%27s_Last_Theorem_for_specific_exponents#Application_to_right_triangles

But did you maybe want just a hint instead??

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If $b\neq 0$, yes, it's true. It follows from the fact that $$a^4-b^4=c^2$$ has no integer solutions, under the assumptions $\gcd(a,b)=1$ and $b\neq 0$. On the other hand, if $c,d$ are two coprime integers such that $c^2+d^2$ is a square, say $e^2$, than $e$ is the sum of two coprime squares. This is the key step of the proof.

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