Determine all value of $p,q\in\mathbb{N}$ such that : $2^{5}5^{3}=(p+1)(2q+p)$

Question

Problem :

Determine all value of $p,q\in\mathbb{N}$ such that :

$$2^{4}5^{3}=(p+1)(2q+p)$$

My try :

$$2q+p-p-1=2q$$

So $2p+q$ odd or $p+1$ odd

I'm going to try all divisible :

$$1,4,5,8,25,125,10,50,250,20,200,500,40,200,1000,16,80,400,2000$$

So : we see that :

$p=0,q=1000$ is a solution

$p+1=16$ and $2q+p=125$ we find $(p,q)=(15,55)$

Also : $250=p+1$ and $2q+p=8$

Is my solution correct?

Answer 1

You are right that one of $p+1$ and $2q+p$ is odd and one even. Furthermore, $2q+p>p+1$.

So $(p+1,2q+p)$ is one of $(1,16\times 125),(5,16\times 25),(25,16\times 5),(16,125)$

The first of these gives $p=0$ which is not a natural number.

Otherwise we have $p=4,q=198$ or $p=24,q=28$ or $p=15,q=55$.

Answer 2

Observe that $p+1 \mid 2^{4}5^{3}$, so we can write $p+1 = 2^{a}5^{b}$, where $a$ and $b$ are integers such that $0\le a\le 4$, $0\le b\le 3$. Now, plug this into the original equation, we then have $2q+p = 2^{4-a}5^{3-b} \Rightarrow 2q = 2^{4-a}5^{3-b} - p \Rightarrow 2q = 2^{4-a}5^{3-b} - 2^a5^b +1$. Now observe that if neither $4-a$ or $a$ were zero, then $2q$ will become an odd number, a contradiction. Thus we must have either one of $4-a$ or $a$ that is equal to zero. Checking the two cases, we have the solutions $(p,q) = (15,55), (4,198), (24,28)$.