Let $a,b,c,d$ be positive integers such that $a^2+3ab+b^2=c^2+3cd+d^2$. Prove that $a+b+c+d$ is a composite number.
I think this statement is correct, at least using the program I came across only composite numbers. An approximate solution plan is as follows: The equality $(2a+3b)^2-5b^2=(2c+d)^2-5d^2$ holds. We can factor it in the ring $\mathbb Z[\sqrt5]$. It is not a UFD, but it is reasonable to assume that there is a further factorization in a wider area - involving what Kummer called "ideal numbers". From here, maybe some information about the coefficients can be extracted. Any ideas?
Suppose $p:= a+b+c+d$ is a prime. Since $d=p-a-b-c$ we have:
$$a^2+3ab+b^2 = c^2+3pc-3ac-3bc-3c^2+p^2-2p(a+b+c)+a^2+b^2+c^2+2ab+2bc+2ca$$ i.e.
$$ab+ ac+bc+c^2=p^2-p(2a+2b-c)$$
or $$(a+c)(b+c) = p(p-2a-2b+c)$$
Last equation implies $p\mid a+c$ or $p\mid b+c$. In each case we have a contradiction, since $p$ is greater then $a+c$ and $b+c$.