Find all integers n for which √(25/2 + √(625/4 –n)) + √(25/2 - √(625/4 –n)) is an integer.

Start by squaring everything: (√(25/2 + √(625/4 -n)) + √(25/2 - √(625/4 -n)))^2 = 25+2sqrt(n)

Since we want integer solutions: (√(25/2 + √(625/4 -n)) + √(25/2 - √(625/4-n)))^2 =x^2, n=y^2

Then x^2 = 25+2y: true for all odd x (positive and negative) and y = -12,-8, 0,12,28,48,72,100,...

So n=0, 144, 784, 2304, 5184, 10000,... (2k)^2(k+5)^2. However, if k=-1, then x is not an integer; it is √41. This corresponds to the case where y=-8. (that y = -12 is a valid solution can be confirmed by substitution).

Edited on March 11, 2015, 2:08 am

Posted by broll on 2015-03-11 01:16:35