Prove that $a+b+c+d$ is a composite number 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?
 A: This issue was discussed there. http://math.hashcode.ru/questions/249507#249550
Similar questions have arisen for many similar equations.
There is a standard approach for solving such problems. And since quite a lot of tasks have been solved using this approach... that is, I think it is already possible to talk about a new direction.
And so that there was no confusion in the approach, I decided to call it holistic algebra. To divide the solution methodology, for example, by algebraic geometry... well, the holistic algebra approach.
Take the equation.
$$X^2+aXY+Y^2=Z^2+aZV+V^2$$
Two solutions can be written. One when mutually simple....
$$X=(a+1)k^2+2tk-t^2-(a+2)qk+q^2$$
$$Y=(a+1)t^2+2tk-k^2-(a+2)qt+q^2$$
$$Z=k^2+akt+t^2-q^2$$
$$V=k^2+akt+t^2+(a+1)q^2-(a+2)(t+k)q$$
The sum of them has the form.
$$X+Y+Z+V=(a+2)(t+k-q)^2$$
And when not...
$$X=k^2+akt+t^2-q^2$$
$$Y=q(aq-at-2k)$$
$$Z=t(aq-at-2k)$$
$$V=t^2-k^2-q^2+aqk$$
$$X+Y+Z+V=(a-2)(q+t)(q+k-t)$$
Well, then everything is quite simple... mutually simple when some brackets are equal to 1... well, or -1
Such a holistic approach allows us to consider any similar equations. Get parameterization. And explicitly write a formula for what their sum looks like.
P.S. --- I asked a question, but they probably didn't understand it. Although it seems to me that the phenomenon has a systemic character.
This is not the first case when there is no regularity from the point of view of algebraic geometry. There is no way to formally describe the phenomenon and write out a formula. On the other hand, the use of holistic algebra gives a pattern.
From the point of view of holistic algebra, there is a pattern, but from the point of view of logic and common sense, there can be no pattern. These are the contradictions that arise all the time.
Let's consider such an equation.
$$X^2+5XY+Y^2=Z^2+5ZV+V^2$$
Let's use the first formula. For example, such coefficients and solutions.
$$(k,t,q) --- (9 ; 5; -3) --- (7 ; 2; -2)$$
$$(X,Y,Z,V) --- (7*107 ; 7*39 ; 7*46 ; 7*97) --- (7*60 ; 7*5 ; 7*17 ; 7*39)$$
When we reduce it by 7. We will get solutions that are not multiples $(a+2)=7$  ....
And it turns out quite funny. On the one hand, the formula speaks of a pattern. It is there. On the other hand, according to the formula, it is possible to obtain such solutions in which this pattern is not observed.
A: 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 $b+c$.
A: November 14.
From a few sources.   For $a^2 + 3ab + b^2 = c^2 + 3cd + d^2$
I believe we get all quadruples with $a,b$ coprime and $c,d$ coprime  by using Gauss composition (Dirichlet version), pretend the common value is compositite  and write as $(p^2 + 3pq + q^2 ) ( r^2 + 3rs + s^2), $ where we don't necessarily require $p,q,r,s$ positive.   The order(p,q; r,s)  gives
$$  a = pr - qs, \; \; \; b = ps + 3sq+qr , $$
while   order(p,q; s,r) gives
$$ c= ps - qr, \; \; \;  d = pr +3rq + qs . $$
$$
\left(
\begin{array}{rrrr}
1&-1&0&0 \\
0&3&1&1 \\
0&0&1&-1 \\
1&1&0&3 \\
\end{array}
\right)
\left(
\begin{array}{c}
pr \\
qs \\
ps \\
qr \\
\end{array}
\right) =
\left(
\begin{array}{c}
a \\
b \\
c \\
d \\
\end{array}
\right)
$$
