0
$\begingroup$

A natural number is said to be congruent if it is the area of a right triangle with rational sides. I've been told that Fermat actually proved his last theorem for $n=4$ by proving that number 1 is not congruent, but I can't seem to find the connection!

It is probably very easy, thanks in advance.

$\endgroup$
1
$\begingroup$

Assume that $n=1$ is congruent. Then there is a right triangle with rational sides $a,b,c$ such that $$ (1)\quad a^2+b^2=c^2, $$

$$ (2) \quad 2ab=4. $$ Adding $(1)$ and $(2)$ we obtain $(a+b)^2=c^2+4$, substracting them gives $(a-b)^2=c^2-4$. Multiplying the new equations gives $$ (a^2-b^2)^2=c^4-2^4, $$ which is a rational solution of $z^2=x^4-y^4$ with $xyz\neq 0$. Then $x^4+y^4=z^4$ has no integer solution with $xyz\neq 0$. The converse implication goes similarly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.