Question

How many integer solutions are there to the equation
x1 + x2 + 2x3 + x4 + x5 = 72

Answer 1

heh, love these questions. the answer is infinite.

Answer 2

if we mean nonnegative solutions, then it is the coefficient of \(x^{72}\) in\[(1+x+x^2+\cdots )^4 (1+x^2 + x^4 + \cdots )\]\[= (1+x^2+x^4 + \cdots ) (1-x)^{-4}\]Now to find the coefficient of \(x^{72}\) we need to find the coefficients of \(x^{72}\), \(x^{70}\), \(x^{68}\), ... in the expansion of \((1-x)^{-4}\) and add them up. The expansion goes like this:\[(1-x)^{-4}=1+4x + 10x^2 + 20x^3 + 30x^4 + \cdots \]The general coefficient being\[\binom{n+3}{3 } = \frac{(n+1)(n+2)(n+3)}{3!} \]We need to evaluate this expression at \(n= 72, 70, 68, ...,0 \) and add it up. Equivalently, this can be expressed as the sum:\[\sum_{s=0}^{36}\frac{(2s+1)(2s+2)(2s+3)}{3!}\]

Answer 3

Let's get to evaluating it though, which is the frustrating part.\[(2s+1)(2s+2)(2s+3)\]\[= 8s^3 + 24s^2+22 s + 6\]Now use\[\sum s = \frac{s(s+1)}{2}\]\[\sum s^2 = \frac{s(s+1)(2s+1)}{6}\]\[\sum s^3 = \left(\frac{s(s+1)}{2}\right)^2 \]

Answer 4

You'll finally get a huge number like \(658 711\).

Answer 5

@jango_IN_DTOWN you can't directly apply that one