Htykuut

What are the number of integer solutions to
$$x_1 + x_2 + x_3 + 2x_4 + x_5 = 72$$ where $x_1 ge 2, x_2, x_3 ge 1, x_4, x_5 ge 0$?

I understand how to do it if it was no "$2x_4$", but just "$x_4$", then its $C(72,4)$... what about with the 2x_4?

Move $2x_4$ to the other side and solve a separate $4$-variable problem for each possible value of $x_4$. In other words, you’re counting non-negative solutions to

$$x_1+x_2+x_3+x_5=68-2k$$

for $k=0,ldots,34$, and you get

$$sum_{k=0}^{34}binom{71-2k}3=sum_{k=1}^{35}binom{2k+1}3;.$$

This actually isn’t quite as nasty as it may look. If we calculate the first few values of $$a_n=sum_{k=1}^nbinom{2k+1}3;,$$ we get $a_1=1$, $a_2=10$, $a_3=35$, and $a_4=84$, with first differences $9$, $25$, and $49$. That suggests that we’re looking at sums of odd squares, i.e., that

$$begin{align*}
sum_{k=1}^{n+1}binom{2k}3&=sum_{k=1}^n(2k-1)^2\
&=4sum_{k=1}^nk^2-4sum_{k=1}^nk+sum_{k=1}^n1\
&=frac23n(n+1)(2n+1)-2n(n+1)+n\
&=frac13n(4n^2-1);.
end{align*}$$

This can be straightforwardly proved by induction on $n$.

EDIT: Thanks @Brian M. Scott for enlightening me. I was confusing two different things.

First up, the different conditions

$$x_1 ge 2; x_2, x_3 ge 1; x_4, x_5 ge 0$$

are rather inconvenient. Let us change the problem to finding the solutions of

$$y_1 + y_2 + y_3 + 2y_4 + y_5 = 68$$ with $y_i ge 0 forall_{i le 5}$

Where we subtracted 2 from $x_1$ and subtracted 1 from each $x_2$ and $x_3$. But then again it is quite inconvenient to have the 2 factor in the middle of the expression. Let us move it to the beginning, shall we? Renaming the variables, we want to solve

$$2x_1 + x_2 + x_3 + x_4 + x_5 = 68, x_i ge 0$$

right? If you fix the value of $x_1 = 2k$, then you want to solve

$$x_2 + x_3 + x_4 + x_5 = 68 - 2k, x_i ge 0$$

right? But that is an easily solvable problem. Since the number of solutions of that equation is given by $C(68 - 2k, 4)$, your answer is just summing up through all values of $68 - 2k$:

$$sum_{i = 0}^{34} C(68 - 2k, 4)$$

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What are the number of integer solutions to
begin{equation}
x_{1} + x_{2} + x_{3} + 2x_{4} + x_{5} = 72 ?quadmbox{where}quad
x_{1} geq 2,;quad x_{2},, x_{3} geq 1,;quad x_{4},, x_{5} geq 0
label{1}tag{1}
end{equation}

That is given by
begin{align}
&sum_{x_{1} = 2}^{infty} sum_{x_{2} = 1}^{infty}
sum_{x_{3} = 1}^{infty} sum_{x_{4} = 0}^{infty} sum_{x_{5} = 0}^{infty}
bracks{x_{1} + x_{2} + x_{3} + 2x_{4} + x_{5} = 72}
\[5mm] = &
sum_{x_{1} = 0}^{infty} sum_{x_{2} = 0}^{infty}
sum_{x_{3} = 0}^{infty} sum_{x_{4} = 0}^{infty} sum_{x_{5} = 0}^{infty}
bracks{x_{1} + x_{2} + x_{3} + 2x_{4} + x_{5} = 68} =
bracks{z^{68}}bracks{1 over pars{1 - z}^{5}pars{1 + z}}
\[5mm] = &
bbox[#ffe,15px,border:2px dotted navy]{ds{528,990}}
end{align}