# equation $x^4 + y^4 = z^4$

Diophantine equations that are insoluble in $\mathbb{Z}$ may become soluble in finite integral domains. Show that

\begin{equation*} x^4 + y^4 = z^4 \end{equation*}

is soluble (as a congruence) in $\mathbb{Z}_{19}$, but is insoluble in $\mathbb{Z}_{17}$.

This problem is giving me a difficult time. I know that $x^4 + y^4 = z^4$ has no solution but, how do i have to make a start to this problem. Does soluble means that it has solution?

• "Is soluble" means "has solutions". Fermat's last theorem implies that $x^4 + y^4 = z^4$ has no solutions among the integers. That does not mean it doesn't have solutions modulo something. The implication goes "There is some modulo with no solutions" $\implies$ "No integer solution". You've been asked to prove that it doesn't have solutions modulo $17$, so this is in line with FLT. – Arthur May 9 '15 at 9:09
• For instance, $2^4+3^4\equiv 1^4\pmod6$. – Arthur May 9 '15 at 9:16
• Start by making a little table of all the values of $x^4$ modulo 17 and modulo 19. – Gerry Myerson May 9 '15 at 9:18

We know that the equation has not solutions over $\mathbb{Z}\backslash\{0\}$, but this does not imply that it doesn't have any solutions over the field $\mathbb{Z}/p\mathbb{Z}$ without using $0$. You are asked to show that there are no solutions when we take $p=17$, but there are when we take $p=19$.
The group $\left(\mathbb{Z}/17\mathbb{Z}\right)^*$ has order $16$. So any element $(x^4)^4\equiv x^{16}\equiv 1\pmod{17}$. That means that any value of $x^4$ in $\left(\mathbb{Z}/17\mathbb{Z}\right)^*$ must be a solution to $y^4\equiv 1\pmod{17}$. The only solutions are $\left\{\pm 1, \pm 4\right\}$. Now you only have to show that there are no combinations of these numbers that satisfy the equation.
The strategy for the case $p=19$ is just to find a set of elements that satisfy the equation. I would just make a list of the values that $x^4$ can take modulo $19$ and try to fit them together to satisfy the equation.