Let $p$ be a prime and let $$(a,p)=(b,p)=1$$. If $x^2 \equiv (b\; mod p)$ and $$x^2 \equiv (a \; mod p)$$ are not solvable, then prove that $x^2 \equiv (ab \; mod p)$ is solvable.
I know that the statements in the theorem(result) is nothing but definition of quadratic residue
I need a short proof as I don't have any idea how to start .
On
Let us start from scratch, i.e. not using Legendre symbols. All you need to know is that the multiplicative group ${\mathbf F_p}^*$ is cyclic (in general, a finite subgroup of the multiplicative group of a field is cyclic), so the quotient ${\mathbf F_p}^*/(squares)$ is cyclic of order $2$, hence isomorphic to the group {${\pm 1}$} (this contains the definition of the Legendre symbol !). In the latter group, we have $(-1)(-1)=1$, and a simple translation gives what you want.
$x^2\equiv a\bmod p$ means $a$ is a quadratic residue modulo $p$ and similarly, $x^2\equiv b\bmod p$ means $b$ is a quadratic residue modulo $p$. So then, in terms of Legendre symbols, we have
$$\Big(\frac{a}{p}\Big)= \Big(\frac{b}{p}\Big)=-1.$$
We need to show $ab$ is a quadratic residue modulo $p$.
We have $$\Big(\frac{ab}{p}\Big)=\Big(\frac{a}{p}\Big)\Big(\frac{b}{p}\Big)=-1\cdot -1=1,$$ as desired.