<s>[INST] On the value of $a+b$ such that $a^2+b^2+3ab=2000$ where $a,b \in \mathbb{Z}^+$ 
Find the value of $a+b$ such that $a,b \in \mathbb{Z}^+$ such that $a^2+b^2+3ab=2000.$

On the outside, this seems like a very easy problem, but I'm not sure how to do this in a non-messy way. What I have so far is that $$a^2+b^2+3ab=(a+b)^2+ab= 2000\implies a+b = \sqrt{2000-ab}.$$ However, is there any way I can do this without trial and error on these values? Solving the quadratic $(a+b)^2 + ab - 2000=0$ wouldn't help a lot either. How should I proceed from this. Thanks in advance.
 [/INST] One less messy way of doing this is to notice that

*

*if $a^2 + 3ab + b^2$ is even, then both $a, b$ must be even.

*if $a^2 + 3ab + b^2$ is a multiple of $25$, then both $a, b$ are multiples of $5$.

Thus $a^2 + 3ab + b^2 = 2000$ leads to $a = 20x, b = 20y$ with $x^2 + 3xy + y^2 = 5$.
Proof of 1:
We have $0\equiv a^2 + 3ab + b^2 \equiv ab + a + b \equiv (a + 1)(b + 1) + 1\bmod 2$. Therefore $a, b$ must be both even.
Proof of 2:
We have $(a - b)^2 + 5ab \equiv 0 \bmod 25$ and hence $a - b \equiv 0\mod 5$. Thus $5ab \equiv 0\bmod 25$ and one of $a, b$ is multiple of $5$. Hence both are multiples of $5$.
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