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Note $$\left(\frac{a^2}{5}\right)$$ is the Legendre symbol.

I used wolfram alpha to see if, $$\left(\frac{a^2}{5}\right) = 1$$ and this is true for integers from $1$ to $10$ and it is except $5$ and $10$, which are divisible by $5$. So is this true in general for any $a$ not divisible by 5?

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    $\begingroup$ @zaira That is simply the definition of the symbol. $\endgroup$
    – egglog
    May 1, 2022 at 17:22
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    $\begingroup$ If you base it off the definition $\Big(\frac{a}{p}\Big) = a^{\frac{p-1}{2}}$ mod $p$, then the result follows immediately from Fermat's Little Theorem. $\endgroup$
    – egglog
    May 1, 2022 at 17:23
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    $\begingroup$ The legendre symbol is multiplicative in the upper entry, so yes! $\endgroup$
    – Lubin
    May 1, 2022 at 17:29

2 Answers 2

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The expression $$\left(\frac{a^2}{5}\right)$$ asks if there exists a natural number $x$ such that $$x^2\equiv a^2 \bmod 5,$$ from the definition of the Legendre Symbol. You simply let $x=a$ to satisfy the equation, so such an $x$ does exist, that is why Wolfram-Alpha returned a $1$ for this Legendre Symbol.

When $5 \mid x$, it is clear that $5 \mid a^2$, so the Legendre gives $0$ by definition.

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Euler's criterion tells us that the Legendre symbol is $\big(\dfrac {a^2}5\big)=(a^2)^\frac{5-1}2\equiv (a^2)^2\equiv a^4\equiv 1\pmod 5,$ by Fermat's little theorem.

This is true whenever $(a,5)=1.$

Thus the answer is yes.

Alternatively $\big(\dfrac {a^2}5\big) =\big(\dfrac a5\big) ^2=1.$

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