Find the number of positive integral solutions to $2a+3b+4c+5d=25$

Question

The number of positive integral solutions of $$2a+3b+4c+5d=25$$

I started solving the equation with the help of the multinomial theorem and but the expression was very long. Are there any alternate ways to solve the problem?

Answer 1

In the product $$(x^2+x^4+x^6+\cdots)(x^3+x^6+x^9+\cdots)(x^4+x^8+x^{12}+\cdots)(x^5+x^{10}+x^{15}\cdots)$$ the coefficient of $x^{25}$ is the number that you are seeking. But this product is $$\frac{x^2}{1-x^2}\frac{x^3}{1-x^3}\frac{x^4}{1-x^4}\frac{x^5}{1-x^5}=\frac{x^{14}}{1-x^2-x^3-x^4+x^6+2x^7+x^8-x^{10}-x^{11}-x^{12}+x^{14}}$$ Now do polynomial division to write enough terms of the inverse: $$x^{14}+x^{16}+x^{17}+ 2 x^{18} + 2 x^{19} + 3 x^{20} + 3 x^{21} + 5 x^{22} + 5 x^{23} + 7 x^{24} + 7 x^{25} + \cdots$$ So there are seven solutions.

Answer 2

This is small enough to allow casework. We know that $a,b,c,d$ are positive, so let $v,w,x,y$ be $a-1,b-1,c-1,d - 1$. Then, we want the non-negative solutions to $$2v+3w+4x+5y=11$$

We now do some casework (not as bad as it looks)

If $y=2$, then there are clearly no solutions.

If $y=1$, then there are three solutions $v,x=1,w=0$, $v,x=0,w=2$ and $v=3,w,x=0$.

So, we now can ignore the $y$ term.

$$2v+3w+4x=11$$

If $x=2$, there is $1$ solution ($w=1,v=0$)

If $x=1$, then there is $1$ solution ($w=1,v=2$)

Now, we can ignore the $x$ term.

$$2v+3w=11$$

If $w=3$, there is $1$ solution ($v=1$)

If $w=2$, there are no solutions

If $w=1$, there is $1$ solution ($v=4$)

If $w=0$, there are no solutions.

So, we are done. We have a total of $7$ solutions!