Determine the total number of ways in which 2⁵*3⁷*5¹¹*11¹³ is expressible as the difference of two nonzero perfect squares.

Extra challenge: A non computer program aided solution.

let n=2^5*3^7*5^11*11^13
then we want integers x>y such that
x^2-y^2=n
(x-y)(x+y)=n
so let n=pq with 0<p<=q
then we have
x-y=p
x+y=q
thus
x=(q+p)/2
y=(q-p)/2
so we need q+p and q-p to both be even in order for x,y to be integers.  Thus p,q are either both even or both odd.  But since they are a factorization of n which is even then at least 1 of p,q must be even.  Thus both p,q must be even.

So we can count all the solutions by counting all the ways of factoring n into a product of 2 even integers.  For each of these factorizations we get a unique solution by setting p equal to the smaller factor and q equal to the larger factor.  Since both factors must be even they must each contain at least 1 factor of 2, thus we can split the 2^5 between them in only two ways namely:
2^1,2^4 and 2^2,2^3
for each of these we can then split the remaining prime factors in any manor which leads to an additional 8*12*14=1344 solutions for each of these.  Thus there are 2*1344=2688 ways to express n as the difference of two nonzero perfect squares.

Posted by Daniel on 2013-09-23 11:14:09