Find all integers n for which
√(25/2 + √(625/4 –n)) + √(25/2  √(625/4 –n)) is an integer.
The domain of f(n)=√(25/2 + √(625/4 –n)) + √(25/2  √(625/4 –n))
Is 0≤n≤625/4=156.25
f is increasing on this domain and f(0)=5 and f(156.25)=√50
So there can be at most 3 solutions and its easy enough to find that there are only two using brute force
f(0)=5,
f(30.25)=6 (n is not an integer)
f(144)=7
but an interesting result on squaring
(f(n))^2 = 25+2√n
This must be a perfect square. Clearly if n=0 it's 25
it can't be 36 and if n=144 it's 49. Beyond that at we've exceeded the domain.

Posted by Jer
on 20150310 12:07:14 