Find all positive integers n which make the expression below an integer:

(25/2 + (625/4 - n)^½ )^½ + (25/2 - (625/4 - n)^½ )^½

(In reply to Simplifying matters (Spoiler) by Steve Herman)

ok first I will show how to compute the square root of a complex number.

sqrt(x+yi)=a+bi

x+yi=(a^2-b^2)+2abi

a^2-b^2=x

2ab=y

b=y/(2a)

b^2=y^2/(4a^2)

a^2-(y^2/(4a^2))=x

4a^4-y^2=4xa^2

4a^4-4xa^2-y^2=0

let t=a^2

4t^2-4xt-y^2=0

t=(4x+-sqrt(16x^2+16y^2))/8

t=(4x+-4sqrt(x^2+y^2))/8

t=(x+-sqrt(x^2+y^2))/2

we want positive t to give real x so we have

t=(x+sqrt(x^2+y^2))/2

this gives

a=sqrt((x+sqrt(x^2+y^2))/2)

and this gives

b=y/(2a)

so we have already determined the solutions for n>=0 and n<=625/4

if n<0 then the first square root is real and the second is complex and thus we end up with a complex number overall and thus never an integer.

if on the other hand n>625/4 then sqrt(625/4 - n) is imaginary which then results in the sum of the square roots of 2 complex numbers.  For the first one we have x=25/2 and y=sqrt(n-(625/4)) and the second has x=25/2 and y=-sqrt(n-(625/4))

doing the substitutions and simplifying we end up with the sum being

(25+2sqrt(n))/2

now this can only be an integer when n is a perfect square.  But then 2*sqrt(n) is always even and thus 25+2sqrt(n) is always odd and thus finally (25+2sqrt(n))/2 can never be an integer.

Now it appears that this proves that n=144 is the only solution but having said that I can not find where the flaw is in Steve's solution and neither can I find a flaw in my own.  So hopefully somebody can find one in either or maybe even both and set the record straight here :-)

Posted by Daniel on 2009-03-24 14:36:47