<s>[INST] $a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. 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. 
 [/INST] If $ \{ a, b \} = \{c, d \}$ then $a+b+c+d = 2(a+b)$ is not a prime.    
Henceforth, we have distinct pairs. If $ a = c$, then $b \neq d$ are roots to $x^2 + ax + a^2 = 0$, so $b+d = -a$ contradicting the requirement that all of the terms are positive.
WLOG, $ a > c \geq d > b$.   
Observe that $$ -(a-b+c-d)(a-b-c+d) = -\left((a-b)^2 - (c-d)^2\right) = 3(ab-cd) = 3\left( (a+b)^2 - (c+d)^2\right) = 3 (a+b-c-d)(a+b+c+d).$$
Since $ a+b+c+d > a-b+c-d \geq a-b-c+d > 0$, we can rewrite the above as a product of positive integers, 
$$  (a-b+c-d)(a-b-c+d)= 3 (c+d-a-b)(a+b+c+d)$$
If $a+b+c+d$ is prime, then the LHS can never be a multiple of $a+b+c+d$ since both terms are smaller, hence we have a contradiction. So $ a+b+c+d$ is composite. 

Notes:    


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*This problem/solution is very reminiscent of 2001/6 IMO. There we had $a^2+ab+b^2 = c^2 -cd+d^2$.    

*I do not know a direct way to show that $cd > ab$. Any thoughts? This seems obvious.   

*This being a contest problem suggests that there is a non "abstract algebra approach", even though it is very tempting. E.g. The equation in my solution can also be derived from WhatsUp's characterization (which is a stronger requirement). 

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