<s>[INST] Determine all value of $p,q\in\mathbb{N}$ such that : $2^{5}5^{3}=(p+1)(2q+p)$ 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?
 [/INST] 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 $(p,q) = (15,55), (4,198), (24,28)$.
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