Expert: Abe Mantell - 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.

I would appreciate any help given.
Thankyou.


Answer
Hello Ben,

Since 2n+1 and 3n+1 are perfect squares, they can be written as:
2n+1=a^2 and 3n+1=b^2, where a, b and n are integers

2n+1 is odd, so a^2 is odd.  Thus, a is odd and can be written as
a=2k+1, with k an integer  Therefore, 2n+1=(2k+1)^2
==> 2n+1=4k^2+4k+1 ==> n=4k^2+4k=4k(k+1), which tells us that
n is a multiple of 4 and must be even.

Also, notice b^2-a^2=(3n+1)-(2n+1)=n, so n=b^2-a^2
Since n is even and a^2 is odd, b^2 must also be odd (since odd-odd=even)
Thus, b must be odd and we can write b=2m+1.

Hence, n=b^2-a^2=(2m+1)^2-(2k+1)^2
n=4m^2+4m+1-4k^2-4k-1
n=4(m^2-k^2)+4(m-k)
n=4(m-k)(m+k)+4(m-k)
n=4(m-k)[m+k+1]

If both m and k are even then so is m-k, then m-k=2N, with N an integer
so n=4*2N(m+k+1)=8N(m+k+1), which is a multiple of 8.

If both m and k are odd, then m-k is even, and the rest follows as above.

If m and k are opposite in parity (i.e. one even and one odd), then
[m+k+1] is even, and the rest follows as above.

Thus, n must be a multiple of 8.  QED

Pretty cool, eh?!!  ;-)

Abe