Answer
The statement $(a+b)^{2}=a^{2}+b^{2}$ is true for some integers but false for others.
Work Step by Step
The simple case $a=b=0$ shows that the statement is sometimes true, since $(0+0)^{2}=0=0^{2}+0^{2}$. However, cases like $a=1$ and $b=2$ falsify the statement, since $(1+2)^{2}=3^{2}=9$, but $1^{2}+2^{2}=1+4=5$.
In fact, since the distributive property (with some simplification) tells us that $(a+b)^{2}=a^{2}+2ab+b^{2}$, the given statement holds if and only if $a^{2}+b^{2}$$=a^{2}+2ab+b^{2}$, i.e., if $2ab=0$, so that $a=0$ or $b=0$.
