Expert: Paul Klarreich - 7/10/2009

Question
Hello there,

My question is:
A positive integer n is such that numbers 2n+1 and 3n+1 are perfect squares. Prove that n is divisible by 8.

What I've tried: Well, I'm in Year 8 and really have no idea how to solve this problem.

I would appreciate any help given.
Thankyou.


Answer
Questioner:   Ben
Country:  Australia
Category:  Advanced Math
Private:  No
 
Subject:  Maths Problem (random, not from a particular topic)
Question:  Hello there,

My question is:
A positive integer n is such that numbers 2n+1 and 3n+1 are perfect squares. Prove that n is divisible by 8.

What I've tried: Well, I'm in Year 8 and really have no idea how to solve this problem.

I would appreciate any help given.
Thank you.
...........................
Hi, Ben,

This is a difficult problem for someone who is only 7 years old, but I'll see what I can do.
------------ PART 1 ----------------
Observation: The odd perfect squares are 1, 9, 25, 49,... each is one more than a multiple of 8.  Let's make that a

Theorem: An odd perfect square can be written as 8M + 1.
Proof:

An odd perfect square is the square of an odd integer:
(2k + 1)^2 = 4k^2 + 4k + 1
= 4(k^2 + k) + 1
= 4k(k + 1) + 1

Now either k or k+1 must be even, so  k(k+1) is even, and can be written 2M:

4[ k(k+1 ] + 1 = 4(2M) + 1 = 8M + 1.
------------- PART 2 -------------
Claim: Your n must be even.

2n + 1 is always odd, and if a P.S. then

2n + 1 = 8M + 1
2n = 8M
n = 4M, so n is even.
------------- PART 3 --------------
So  n is even.  Then  2n+1 and 3n+1 are both odd P.S.'s
Therefore  2n+1 = 8M + 1 and
          3n+1 = 8L + 1
and we subtract:
n = 8(L - M), which is divisible by 8.

FINITO.