<s>[INST] Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable 
Prove that if $b^2-4ac=k^2$ (for some positive integer $k$ ) then $ax^2+bx+c$ is factorizable

I related this to the roots of equation: $ax^2+bx+c=0$ and using roots $x_1,x_2$:
$(x-x_1)(x-x_2)=0\implies(2ax+k-b)(2ax+k+b)=0\implies 4a^2x^2+4akx+k^2-b^2=0$ but this seems has nothing to do with original equation.Any better way??!!  
Under what conditions can we be sure the factorization involves just integers?? 
 [/INST] $$ax^2+bx+c=a\left(\left(x-\frac{b}{a}\right)^2-\frac{b^2-4ac}{a^2}\right)=a\left(\left(x-\frac{b}{a}\right)^2-\frac{k^2}{a^2}\right)$$
which is of the form $$\alpha ^2-\beta^2=(\alpha -\beta )(\alpha +\beta ).$$
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