<s>[INST] If $a^2+ab+b^2=c^2+cd+d^2$ then $a+b+c+d$ is not prime. Let $a,b,c,d$ be positive integers such that $a^2+ab+b^2=c^2+cd+d^2$. Show that $a+b+c+d$ is not prime.
My proof looks like this:
$(a+b)^2 - ab=(c+d)^2-cd$
$(a+b)^2 - (c+d)^2=ab-cd$
$(a+b+c+d)(a+b-c-d)=ab-cd$
$a+b+c+d=\frac{ab-cd}{a+b-c-d}$
I'd like to have $a+b+c+d$ as product (of integers) not quotient
 [/INST] It is necessary to use the formula.   Diophantine equation $a^2+b^2=c^2+d^2$ 
For example this.  
You can write a similar equation and solutions:  $$a^2+ac+c^2=x^2+xy+y^2$$  
Solutions have the form:  $$a=q^2+k^2-p^2+kq$$  
$$c=q^2+k^2+2p^2+kq-3pk-3pq$$  $$x=q^2-2k^2-p^2+3pk-2qk$$  $$y=k^2-2q^2-p^2+3pq-2qk$$
This means that:
$$a+c+x+y=q^2+k^2-p^2-2qk=(q-k-p)(q-k+p)$$
Easy number can always choose.
$$(q-k-p)(q-k+p)=1*19$$
Then for example: it turns out one of the numbers is negative.  If we change the signs. $x=-x$ ; $y=-y$ . You get a multiple of 3 and the square. So simple number can be when $q=p-k+1$
This means that at least one number must be again negative.
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