Determine if there is a solution to the congruence $x^6+5x^2 \equiv 2\mod 7$

Question

I've checked if $x = 1, 2, \ldots, 6$ and they are not congruent to $2 \mod 7$ so, $x$ would have to be a fraction if anything. How would I be able to show it?

Answer 1

If you've checked $x = 0, 1, 2,3,4,5, 6$ and have found that $x^6+5x^2$ is never congruent to $2 \bmod 7$, then there is nothing else to prove.

Answer 2

First observe that $x\equiv 0\mod 7$ is not a solution. So we may suppose $x$ is not divisible by $7$, and by lil' Fermat, we know that $x^6\equiv 1\mod 7$. Therefore, we obtain the congruence $$5x^2\equiv 1\mod 7 \iff x^2\equiv 3\mod 7,$$ since the inverse of $5\bmod 7$ is $3$.

What is the list of non-zero squares mod. $7$, knowing the non-zero elements are $\pm 1,\pm 2,\pm 3$?

Answer 3

Apparently you are already done.

But we have $5x^2\cong1\pmod 7$ by Fermat's little theorem, if $x\neq0$. A quick check reveals:

$0: 0^6+ 5\cdot 0^2\cong0\not\cong 2\\1:5\cdot1^2\cong5\not\cong1\\2:5\cdot 2^2\cong20\cong6\not\cong1\\3:5\cdot3^2\cong45\cong3\not\cong1\\4:5\cdot 4^2\cong80\cong 3\not\cong1\\5:5\cdot5^2\cong125\cong6\not\cong1\\6:5\cdot 6^2\cong180\cong5\not\cong1$, all $\pmod7$.

Answer 4

$\bmod 7\!:\ x\not\equiv 0\,\Rightarrow\, \color{#c00}{x^{\large 6}\equiv 1}\,$ so it is $\,1\equiv -2x^{\large 2}\,\overset{\large (\ \ )^{\LARGE 3}}\Longrightarrow\,1\equiv -\color{#c00}{x^{\large 6}}\equiv -\color{#c00}1\,\Rightarrow\!\Leftarrow$