Select Page

# 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
Number Theory
5/10
##### Suggested Book
Challenges and Thrills of Pre College Mathematics

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.

# 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.

# Similar Problems

## Tetrahedron Problem | AIME I, 1992 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron Problem.

## Triangle and integers | AIME I, 1995 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.

## Functional Equation Problem from SMO, 2018 – Question 35

Try this problem from Singapore Mathematics Olympiad, SMO, 2018 based on Functional Equation. You may use sequential hints if required.

## Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and the greatest integer.

## Arithmetic sequence | AMC 10A, 2015 | Problem 7

Try this beautiful problem from Algebra: Arithmetic sequence from AMC 10A, 2015, Problem. You may use sequential hints to solve the problem.

## Series and sum | AIME I, 1999 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.

## Inscribed circle and perimeter | AIME I, 1999 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Rectangles and sides.

## Problem based on Cylinder | AMC 10A, 2015 | Question 9

Try this beautiful problem from Mensuration: Problem based on Cylinder from AMC 10A, 2015. You may use sequential hints to solve the problem.

## Median of numbers | AMC-10A, 2020 | Problem 11

Try this beautiful problem from Geometry based on Median of numbers from AMC 10A, 2020. You may use sequential hints to solve the problem.

## LCM and Integers | AIME I, 1998 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998, Problem 1, based on LCM and Integers.