Understand the problem

Let $p,q$ be prime numbers$.$ Prove that if $p+q^2$ is a perfect square$,$ then $p^2+q^n$ is not a perfect square for any positive integer $n.$
Source of the problem
Italy MO 2019, Problem 2
Topic
Number Theory
Difficulty Level
5/10
Suggested Book
Challenges and Thrills of Pre College Mathematics

Start with hints

Do you really need a hint? Try it first!

Let us first write the condition and use the idea that p and q are prime. Let $a \in \mathbb{N}$ such that \[ p + q^2 = a^2 \] =>   \[ p = a^2 - q^2 = (a - q)(a + q) \]. As $p$ is a prime and $a + q > a - q$, then $a - q = 1$. We must have $p = 2q + 1$. Now, try to use this condition to rewrite the expression we are interested in.
p = 2q+1. Suppose that there exists $b \in \mathbb{N}$ such that for some positive integer $n$, then
\[ p^2 + q^n = b^2 \] => \[ q^n = b^2 - p^2 \] => \[ q^n = (b - p)(b + p) \]. It implies that both (b-p) and (b+p) must be powers of q. Hence gcd(b-p, b+p) = (b-p). b-p | b+p => b-p | 2p = 4q + 2. As, b-p is a power of q, it implies that b-p = 1 and q can be a prime or q = 2 and b-p = 2 as gcd (b-p, 4q+2) = 1 if q is an odd prime or 2 if q = 2. Thus, we get that ( b-p = 1 ) or ( b-p =2, q = 2 ) and p = 2q+1.  
Case 1 ( b-p = 1 ): b -p = 1,  (b – p) = 1, (b+p) = \(q^n = 1 + 2p = 3 + 4q\). Now, you see that LHS will grow much more fast as q increases than the RHS.  Mathematically, \( q^n \geq q^2 \geq 4q+3 \) as \( q \geq 5, n \geq 2 \). So, the only possibility for q is q = 2, 3, 5. But it has no solution as you can see.
Case 2 ( q = 2 ):  It implies p = 2q+1 = 5. \( 2^n = (b-5)(b+5) \). It implies that the both b-5 and b+5 must be powers of 2.  Let \( b-5 = 2^a \leq b+5 = 2^b \rightarrow 2^b – 2^a = 10 \rightarrow a = 1 and 2^(b-1) – 1 = 5 \). This has no solution. 

QED.

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