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I have a equation and I need to find the number of possible integer solutions for it. The equation is of the form $$2a+2b+2c+d=n, (a,b,c,d > 0)$$

UserX
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Vivek Maskara
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    Have you tried searching. This question or one closely similar has come up several times in the last week. – almagest Sep 11 '14 at 14:04
  • Yes, I tried but I couldn't get a solution except for the general case with no restrictions – Vivek Maskara Sep 11 '14 at 14:09
  • See also http://math.stackexchange.com/questions/602547/how-many-positive-integer-solutions-does-the-equation-x-y-z-w-15-have?rq=1, or http://math.stackexchange.com/questions/657515/how-many-solution-of-a-equations?rq=1. – Dietrich Burde Sep 11 '14 at 14:10
  • It boils down to $\sum_k\binom{2k+2}{2}$ which should give a cubic. It's pretty straightforward but perhaps there are more clever solutions. – Bart Michels Sep 11 '14 at 14:12
  • @VivekMaskara The restriction to positive integers is trivially dealt with. Anyway the idea is to help you solve things yourself, not provide solution service! :) – almagest Sep 11 '14 at 14:13
  • yep positive integers is no longer a problem. @Burde's link helped – Vivek Maskara Sep 11 '14 at 14:15
  • Is there any restriction on $n$? – Hassan Muhammad Sep 11 '14 at 14:19

2 Answers2

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Use the generating function of the LHS:

$$f(x) = (x^2+x^4+x^6+...)^3(x+x^2+x^3+...)$$

and the answer for the equation is the coefficient of $x^n$ in the generating function.

To get the cofficient, we split to disjoint events, by the size of $d$, obviously that if $n$ is even/odd then $d$ is even/odd, and then divide $n-d$ by $2$ and split it to $3$ different cells with repetitions $$CC_3^{\frac{n-d}{2}}$$

Then sum over all those events.

Snufsan
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Idea:

Let $a+b+c=k$, then we have $2k+d=n$. Now for any given value of $d$ there is corresponds $3$ values of $k$. So for $m$ values of $d$ we have $3m$ values of $k$

So that we are going to have ... solutions.

Hassan Muhammad
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