# Understand the problem

Let be prime numbers Prove that if is a perfect square then is not a perfect square for any positive integer

##### 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 such that => . As is a prime and , then . We must have . Now, try to use this condition to rewrite the expression we are interested in.

p = 2q+1. Suppose that there exists such that for some positive integer , then

=> => . 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.

=> => . 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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