# Equivalent statement of Fermat's last theorem

I'd like to know how to show that, if there are no integer solutions to $a^n + b^n = c^n$ for $a, b, c, n \in Z$ and $n > 2$ then this is equivalent to either $a$ or $b = 0$ are the only rational solutions to $a^n + b^n = 1$ for $n > 2$. Not sure if this is a simple proof or not, could someone provide some pointers?

• Note that $a=b= 0$ is not a solution to $a^n+b^n=1$ – M Turgeon May 30 '12 at 14:31
• Hint: an equation like this in $\mathbb Q$ can be multiplied by something to turn it into an equation in $\mathbb Z$. – Karolis Juodelė May 30 '12 at 14:34
• Isn't this identical to Fermat's Last Theorem? Or am I missing something? – user26649 May 30 '12 at 14:58
• @Farhad: you are not. The OP is. So you (or someone else) may want to give an answer as to how they are equivalent. (Sort of the point of a Q&A website, that is. :p ) – Willie Wong May 30 '12 at 15:19
• You've got the statements slightly wrong. – Arturo Magidin May 30 '12 at 16:17

You got the statements slightly wrong. I believe what you meant to write as the two statements is:

• Statement 1. If $n\gt 2$, and $a,b,c\in\mathbb{Z}$ are integers such that $a^n+b^n=c^n$, then one of $a$, $b$, and $c$ are equal to $0$. (There are no nonzero integer solutions to $a^n+b^n = c^n$ if $n\gt 2$).

• Statement 2. If $n\gt 2$, then the only rational solutions to $a^n+b^n = 1$ have $a=0$ or $b=0$.

Statement 1*$\implies$*Statement 2

Suppose $a$ and $b$ are rational numbers such that $a^n+b^n = 1$. Finding a common denominator, we can write $a=\frac{c}{m}$ and $b=\frac{d}{m}$, with $c,d$ integers, $m\gt 0$. Therefore, we have $$\frac{c^n}{m^n} + \frac{d^n}{m^n} = 1.$$ Multiplying through by $m^n$ we obtain $$c^n + d^n = m^n$$ with $c,d,m$ integers, $n\gt 2$. By Statement 1, at least one of $c$, $d$, and $m$ must be equal to $0$. We know that $m$ is not equal to $0$, which means either $c=0$ or $d=0$. If $c=0$, then $a=\frac{c}{m} = 0$; if $d=0$, then $b = \frac{d}{m} = 0$. So we conclude that if Statement 1 holds, and $a^n + b^n = 1$ with $a$ and $b$ rationals, $n\gt 2$, then either $a=0$ or $b=0$, as claimed.

Statement 2*$\implies$*Statement 1

Let $n\gt 2$ and let $a,b,c$ be integers such that $a^n+b^n=c^n$. We want to show that at least one of $a$, $b$, and $c$ are equal to $0$. If $c=0$, then we are done. If $c\neq 0$, then dividing through by $c^n$ we have $$\frac{a^n}{c^n} + \frac{b^n}{c^n} = 1$$ or $$\left(\frac {a}{c}\right)^n + \left(\frac{b}{c}\right)^n = 1.$$ By Statement 2, we must have $\frac{a}{c}=0$ or $\frac{b}{c}=0$. In the first case, we have $a=0$; in the second, we have $b=0$. So, either $a=0$, $b=0$, or $c=0$, as claimed. $\Box$

Note that there are certainly integer solutions to $a^n+b^n=c^n$: take $a=c=1$, $b=0$. The point is that there are no nonzero solutions.