Determine all possible values of a
nonnegative integer n such that;
• 2n-1*n +1 is a perfect square.
Note: Adapted from a problem which appeared in a Georgia IMO.
The perfect square must be odd, so we can write this as an equation
2^(n-1)*n+1 = (2k+1)^2
Then simplify
2^(n-3) * n = k * (k+1)
k and k+1 are consecutive integers, so exactly one is even. That factor (k or k+1) must then be a multiple of 2^(n-3).
If for any n we have 2^(n-3) > n+1 then the factorization becomes impossible since the power of 2 will be too far apart from any possible factor of n.
If n=6 we have 2^(n-3) = 2^3 = 8 > 7 = n+1. Increasing the value of n will only make the difference more pronounced. Then we conclude that any n>=6 will have 2^(n-3) > n+1.
But this makes those values of n impossible to have 2^(n-3)*n factor how we want. Therefore n<=5.
Then its just simple arithmetic for n=0 to 5 to find n=0 -> 2^(-1)*0+1 = 1^2 and n=5 -> 2*4*5+1 = 9^2 are the only n which yield a perfect square. Then all all possible values of a nonnegative integer n are 0 and 5.
Solution
