# 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.

# Watch video

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

Google