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  √(625/4n)))^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 20150311 01:16:35 