Equi distant perfect squares
Number Theory
Level
2
What is the maximum possible value of n such that n is an integer and n-124 and n+124 are perfect squares?
The answer is 3845.
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1 solution
Let n - 124 = k^2 and n + 124 = r^2
Follows that k^2 - r^2 = 248
or (k+r)(k-r) = 248 = 124 * 2 or 62 * 4 or 31 * 2^3 or 248 * 1
The valid integer solutions for k and r are k = 63 and r = 61 giving n = 3845
or k = 33 , r = 29 giving n = 965.
The greater of the two solutions is 3845 and hence the answer to this question is 3845.