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Find a solution or show that there is none: $x^2 \mod 7 = 3\;(x\in \mathbb{Z})$

I've tried different ways to approach this problem, such as rewriting the equation in the form $x^2 = 7k+3$, where k is an integer.

Then it becomes proving that $\sqrt{7k+3}$ is not an integer (for $k\in \mathbb{Z}$), or finding a solution to it. But here, I've been stuck.

Can anybody help me with this?

Thank you for your time.

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    Calculate $0^2,1^2\cdots,6^2$ modulo 7 & observe that $3$ is not on the list. – 2017-02-25
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    How comes this shows that there is no solution? – 2017-02-25
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    @DrC: Because every integer is congruent, mod. 7, to one of theese : 0,1,2,3,4,5 or 6. – 2017-02-25
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    The only candidates solutions are $0, 1, 2, 3, 4, 5, 6$. So just try them all and verify that none is solution. – 2017-02-25

4 Answers 4

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The question of whether the equation $$x^2 \equiv 3 \pmod{7}$$ has any solutions is equivalent to the question is $$x^{2} = 3$$ solvable for $x \in \mathbb{Z}/ 7\mathbb{Z}$.

So, as Donald Splutterwit mentions in the comments, we can calculate the value of $x^{2}$ for each $x \in \mathbb{Z}/ 7\mathbb{Z}$ and check whether it is equal to $3$.

In general, the question of whether for $p$ prime and fixed integer $n$ the equation, $$x^{2} \equiv n \pmod{p}$$ has a solution is answered by Gauss's quadratic reciprocity.

$\textbf{Edit:}$

To expand upon the answer, suppose that there exists $x\in \mathbb{Z}$ such that $$x^{2} \equiv 3 \pmod{7},$$ then $7 \mid (x^{2} - 3)$, which implies that there is an $a\in \mathbb{Z}$ so that $$x^{2} - 3 = 7a.$$

Then, by the Euclidean division algorithm, $x = 7k+r$ for some $0 \leq r \leq 6$. Substituting this expression into the above formula yields $$49x^{2} + 14kr + r^{2} - 3 = 7a,$$ which is equivalent to saying that there is $a^{'} \in \mathbb{Z}$ so that $$r^{2} - 3 = 7a^{'}.$$ (Just bring the terms on the LHS divisible by $7$ to the RHS and factor out a $7$.)

Now, we see that if $7 \mid (x^{2} - 3)$, then $7 \mid (r^{2} - 3)$ for some $0 \leq r \leq 6$.

So, to show that $7 \nmid (x^{2} - 3)$ for all $x\in \mathbb{Z}$, it is enough to show that $7 \nmid (r^{2} - 3)$ for $r \in \{0,1,\ldots, 6\}.$

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    What is Z/7Z? I'm not familiar with this terminology, does it mean multiples of 1/7? – 2017-02-25
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    $\mathbb{Z} / 7\mathbb{Z}$ refers the integers modulo $7$. It can be thought of as a set of equivalence classes. So, you can write that $\mathbb{Z} / 7\mathbb{Z} = \{0 + 7\mathbb{Z}, 1 + 7\mathbb{Z}, \ldots, 6+7\mathbb{Z}\}$ where $0 + 7\mathbb{Z}$ is the set of all integers divisible by $7$, and in general, $n + 7\mathbb{Z} = \{ x\in \mathbb{Z} \; : 7 \mid (x-n)\}.$ In short, $\mathbb{Z} / 7\mathbb{Z}$ is a convenient mathematical structure where the usual rules of modular math apply. – 2017-02-25
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Any $x\in Z$ can be written as

$x = 7k+r$,

$k,r\in Z$ and $r\in [0,7)$

Then $x^2 = 49k^2+2(7kr)+r^2$

All you now have to check is whether, for any possible value of $r^2$, you get $3(mod 10)$.

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Since $7$ is an odd prime and $7$ doesn't divide $3$, we can use Euler's criterion first to check whether the following congruence has a solution:

$$x^2 \equiv 3 \pmod{7}$$

By Euler's criterion: $$ 3^{\tfrac{7-1}{2}} \equiv 27 \equiv 7*4 - 1 \equiv -1 \pmod{7} $$

That means there is no solution at all.

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Hint $\ {\rm mod}\ 7\!:\,\ \left[x^{\large 2}\equiv 3\right]^{\large 3}\!\Rightarrow\, x^{\large 6}\equiv -1\,$ contra little Fermat