For [tex]x^2+mx+m[/tex] to be a perfect square trinomial, we would need to have
[tex](x+k)^2=x^2+2kx+k^2=x^2+mx+m[/tex]
which means [tex]m,k[/tex] must satisfy
[tex]\begin{cases}2k=m\\k^2=m\end{cases}\implies2k=k^2[/tex]
So
[tex]2k=k^2\implies k^2-2k=k(k-2)=0\implies k=0\text{ or }k=2[/tex]
If [tex]k=0[/tex], then [tex]m=0[/tex], but then [tex]x^2+mx+m[/tex] is a monomial. If [tex]k=2[/tex], then [tex]m=4[/tex].
