if $a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$

Question

If $a+b$ is an positive integer and $a\ge b$ and $a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$.

I tried a lot to solve it, i came to a step after which i was not able to proceed forward, the step was $(a+b)^3 - 41\cdot17 (a+b)^2 - 3ab(a+b) - 6ab = 0$ Please help me to solve this.

Answer 1

Note: I solve it base on that a,b are positive numbers(not integers)

suppose $s=a+b$

Then, $a^3+(s-a)^3 +3(a^2+(s-a)^2) - 700s^2 = 0$

Rearrange it, $(6+3s)a^2-(3s^2+6s)a+(s^3-697s^2)=0$

$\Delta=(3s^2+6s)^2-4(6+3s)(s^3-697s^2)=-3s^4+8376s^3+16764s^2>0$

$0<s<=2794$

for one solution, $0<a<s$

(1) $0<3s^2+6s-\sqrt\Delta<2s(6+3s)$

L:$3s^2+6s>\sqrt\Delta$

$(3s^2+6s)^2>(3s^2+6s)^2-4(6+3s)(s^3-697s^2)$

$4(6+3s)(s^3-697s^2)>0$

$0<s<697$

R:$-3s^2-6s<\sqrt\Delta$ It is always true

(2) $0<3s^2+6s+\sqrt\Delta<2s(6+3s)$

L: always true

R: $3s^2+6s>\sqrt\Delta$

It is true for $0<s<697$

Overall $a+b=s=1,2,...695$ or $696$

$sum=242556$

Answer 2

Here is a start, not a full answer.

Let $s=a+b$ and $p=ab$. Then $3 p (s + 2) = (s - 697) s^2$ as WA tells us.

Then $s+2$ divides $(s - 697) s^2$ and so $s+2$ divides $2796 = 2^2 \cdot 3 \cdot 233$.