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Equal differences (Posted on 2015-08-23) Difficulty: 4 of 5
Given the positive integers, x, y, and z, are consecutive terms of an arithmetic progression, the least value of the positive integer, n,
for which the equation, x^2 – y^2 – z^2 = n, has exactly two solutions is n=27
since 34^2 – 27^2 – 20^2 = 27
and 12^2 – 9^2 – 6^2 = 27.
It turns out that n=1155 is the least value for which there are exactly ten solutions.

How many values of n less than one million have exactly ten distinct solutions?
Source: Project Euler

No Solution Yet Submitted by Ady TZIDON    
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a start Comment 1 of 1
Set x=y+d and z=y-d and substitute to get y(4d-y)=n.  By looking at the factors of n we can readily find solutions, such as (40,21,2) and (482,385,288) for n=1155.

For n with exactly 10 solutions we need exactly 10 values for y.  That lets out cases where n=prime or n=product of exactly 2 or 3 primes.  Not all factors are solutions either because z=y-d>0, so that 4y>4d, 3y>(4d-y), and 3y^2>n.  In the  case of n=1155 that means y>19, excluding possible solutions 1,3,5,7,11,15. 

I don't see how to get closer to a full count without a computer program.   

  Posted by xdog on 2015-08-24 12:15:16
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