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My book mentioned the following property of Legendre symbol:

$$\left(\frac{a^2}{p}\right) =1, $$

And it said in the proof That the integer a trivially satisfies the congruence $x^2 \equiv a^2 \pmod{p}$; hence, $\left(\frac{a^2}{p}\right) =1.$

But I do not understand how "a trivially satisfies the congruence $x^2 \equiv a^2 \pmod{p}$", could anyone explain this for me please?

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    $\begingroup$ All that says is that $a^2\equiv a^2\pmod p$ $\endgroup$ – lulu Jun 12 at 10:15
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The legendre symbol

$$\left(\dfrac{k}{p}\right)$$

is defined to be $1$ if there exists $x$ such that $x^2\equiv k\mod p$. When $k=a^2$, we can choose $x=a$ and thus $x^2\equiv a^2\equiv k\mod p$; thus by definition the Legendre symbol is 1.

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Another argument is, that we have $$ \left(\frac{a^2}{p}\right) =\left(\frac{a}{p}\right)^2=(\pm 1)^2=1 $$ with $p\nmid a$, because the Legendre symbol is multiplicative.

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