2

We work in $\mathbb{Z}/11\mathbb{Z}$. I want to find the integers $n$ such that the equation $x^n+y^n=0$ has solutions different from $(0,0)$.

It's obvious for any odd number $n$.

If $n$ is even we can look at what happens only when $x \in \{1,2,3,4,5\}$. We'll see that for those $x$ and for any $n \in${2,4,6,8}, $x^n\in \{1,3,4,5,9\}$.

As $x^{10}=1$ for any $x$, It's easy to conclude that if $n$ is even there is no non trivial solution to $x^n+y^n=0$.

Now my problem is this : I had to see manually what happens for any even $n<10$ which I find non satisfactory. Is there a more generic way to Handle this problem for any prime $p$ by the way in $\mathbb{Z}/p \mathbb{Z}$. ?

138 Aspen
  • 880
Will Tracy
  • 21

2 Answers2

2

Considering $n=2$ is enough to show that there are no solutions for even $n$. Suppose that we know there are no non-trivial solutions for $n = 2$.

If $n = 2k$ and $x^n + y^n = 0$, you get $(x^k)^2 + (y^k)^2 = 0$. Thus, $(x^k, y^k)$ is a solution for the $n = 2$ case. This implies $x^k = y^k = 0$, so $x = y = 0$.

Tzimmo
  • 1,299
1

Partial Hint: Check when 0 can be expressed as sum of two quadratic residues.

For instance: In case of $p=13$, see that $-1$ is a quadratic residue (since $-1\equiv 5^2 \pmod {13}$). Hence, $$5^2+1^2 =0.$$

In case of $p\equiv 3 \pmod 4$, 0 can never be expressed as sum of two quadratic residues. Hence coupled with the previous answer, in case of $p\equiv 3 \pmod 4$, for even $n$ your equation has no non trivial solution.

So you need to only look in $\mathbb{Z}_p$ where $p\equiv 1 \pmod 4$.

Yathi
  • 1,854