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^3-44k^2-48k+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)
25-t^2=2sqrt(n)
(25-t^2)^2=4n
n=(25-t^2)^2/4
25 = 1 mod 4
if t is odd then t^2 = 1 mod 4 and thus 25-t^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^3-44k^2-48k+144
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Posted by Daniel
on 2015-03-10 15:13:01 |
re(2): solution & playing around
