$2\cdot x^n\cdot y^n \equiv -1$ mod $p$ is not true

Question

Show that this congruence can never be true: For every prime $p=2n+1$ , $n$ a positive integer , we can find integers $x, y$ such that:

$$2\,x^n\,y^n \equiv -1 \mod p$$

Answer 1

Take $p=5$ Note for $x,y \in \mathbb{Z}/5\mathbb{Z}$ then $x^2,y^2 \in \{0,\pm 1\}$

$$2x^ny^n= \begin{cases} 0 & \mod 5\\ 2 & \mod 5\\ -2 & \mod 5\\ \end{cases}$$

Since none of the options are equal to $-1 \mod 5$ the stated equation cannot be true for $p=5$

Answer 2

Let $n = 2$, $p = 5$. Then $2 x^2 y^2 \in \{0,2,3\} \pmod{5}$. This is because the quadratic residues modulo $5$ are $\{0, 1, -1\}$ and $2$ multiplied by any sequence of elements of $\{0,\pm 1\}$ is in $\{0,\pm 2\}$.

Answer 3

This seems to never be true, unless $p=3$:

Assume that $2(xy)^n\equiv_p -1$, then $$4(xy)^{2n}\equiv_p 1$$ Since $2n=p-1$ we have $$4\cdot 1\equiv_p 1$$ Which holds when $p=3$.

Answer 4

Your question is quivalent to:

$$\,2(x \cdot y)^{k}+1 \equiv 0 \mod (k+1)$$ Where $k+1$ is prime and $k=2n$

Which would only have solution when:

$$\,(x \cdot y)^{2n} \equiv n \mod p$$

Trivial Solution: When $x \cdot y = 1$ and $n=1$. Hence have a solution when $p=3$

Now let $x \cdot y = a$, so that the following is true:

$$a^{p-1} \equiv n \mod p$$

We have two cases:

a. a is divisible by p: This would imply $a^{p-1} \equiv 0 \mod p$ which is the first contradiction.

b. a is NOT divisible by p: By Fermat's Little Theorem, this would imply $a^{p-1} \equiv 1 \mod p$ which is the second and final contradiction.

Conclusion: Your question is always true for $p>3$. That is $$2\,x^n\,y^n \equiv -1 \mod p$$ is always NOT TRUE when $p>3$