Let k be a positive integer.
Prove that there must exist a positive integer n such that the sets
A( A={1^2, 2^2, 3^2, . . . })
and B ( B = {1^2+n, 2^2+n, 3^2+n, . . . }) share exactly k identical members.
One way to restate this problem is to say that it is possible to choose a positive integer n such that the equation:
a^2 = b^2 + n (a, b also positive integers) has exactly k solutions, regardless of choice of k > 0
Rewriting,
n = a^2  b^2 = (a+b)(ab)
Since b > 0, these two factors must be distinct (if a+b = ab then b must = 0)
Since a and b are integers, these two factors must have the same parity, either both odd or both even. (if one were odd and the other even then their sum would be odd but (a+b)+(ab) = 2a which must be even)
So the number of solutions is equal to the number of ways to factor n into two distinct factors that are either both odd or both even.
You can construct an n with any desired number of solutions. For example n = 3*2^(k+1) has exactly k such factorizations: The 3 must belong to one factor, each factor must have at least one 2, and any number in [0,k1] (k choices) of the remaining 2's can belong to the factor that has the 3. In the case of k = 2, n= 24 is one such desired number. There are exactly 2 solutions to a^2 = b^2 + 24  7^2 = 5^2 + 24 and 5^2 = 1^2 + 24.
Clearly there are an infinite number of n's that server for any k  any prime > 2 could be substituted in my example for "3" and the logic would still hold. Of course only one such n is required for the existence proof the problem requests.

Posted by Paul
on 20170216 11:34:57 