$a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number? Let $a,b,c,d$ be natural numbers such that $a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number?

Now assume that $a+b+c+d = p > 2$ for some choice of $a,b,c,d$. Notice that we cannot have $a=b=c=d$. Also notice that $(a+b+c+d)^{2}$ must be odd and only have three factors: $1, p, p^{2}$.
$$ p^{2} = a^{2} + b^{2} + c^{2} + d^{2} + 2(ab+ac+ad + bc+bd + cd) $$
$$ = 2(c^{2}+d^{2}+ac+ad++bc+bd) + (ab + 3cd) $$
$$ =2(c^{2}+d^{2}+ac+ad++bc+bd + cd) + (ab + cd) $$
So $(ab+cd)$ must be odd.
Now if $a+b+c+d$ is prime $>2$ then either 3 of them is odd and 1 is even, or 3 of them is even and one is odd. 
$WLOG$, let $a,b,c$ be even and $d$ is odd, then $a^{2} + b^{2}+ ab$ is even and $c^{2} + d^{2}+ cd$ is odd, so we can't have 3 even and 1 odd.
But it is possible for 3 odd and 1 even. 

Another solution, whose approach is distinct from the rest.
Suppose that for some prime $ p \geq 4$,  $ a + b + c + d = p$.
Then $ a+ b \equiv - (c+d) \pmod{p}$,
$ a^2 + 2ab + b^2 \equiv c^2 + 2cd + d^2 \pmod{p}$,
$ab \equiv cd \pmod{p}$.
$a^2 - 2ab + b^2 \equiv c^2 - 2cd + d^2 \pmod{p}$
WLOG $ a - b \equiv c - d \pmod{p}$
Hence $ 2a \equiv (a-b) + (a+b) \equiv (c-d)  - (c-d) \equiv - 2d \pmod{p}$.
Since $ p \neq 2$, so $ a + d \equiv 0 \pmod{p}$.
Then, $ a + b + c + d > a + d \geq p $ which is a contradiction.
