Find all integers n for which
√(25/2 + √(625/4 –n)) + √(25/2  √(625/4 –n)) is an integer.
(In reply to
re: solution & playing around by Charlie)
generalizing this result we have that
n=4k^4+8k^344k^248k+144 for integer k>=1
gives all values of n for which f(n) is an integer
this can be derived from the result of:
f(n)^2=25+2sqrt(n)
thus if f(n)=t then
t^2=25+2sqrt(n)
25t^2=2sqrt(n)
(25t^2)^2=4n
n=(25t^2)^2/4
25 = 1 mod 4
if t is odd then t^2 = 1 mod 4 and thus 25t^2 = 0 mod 4
thus we simply need that k be an odd integer
substituting t=2k+1 we get
n=(25(2k+1)^2)^2/4
n=4k^4+8k^344k^248k+144

Posted by Daniel
on 20150310 15:13:01 