Determine all possible positive integer values of n that satisfy this equation:

Σ

j=n j=1

(x+j-1) (x+j) = 10*n

whenever x is an integer.

Note: For a desired value of n, the above equation holds true for at least one value of x.

(In reply to re(2): Possible solution by broll)

this is how I worked it out
sum  (x+j-1)(x+j)
sum j^2+(2x-1)j + x^2-x
[sum j^2] + (2x-1)*[sum j] + [sum x^2-x]
now we know formulas for sum j and sum j^2 thus we get
(1/3) * (n^3+3n^2x+3nx^2-n)=10n
n^3+3n^2x+3nx^2-n=30n
3nx^2+3n^2x+n^3-29n=0
n is not zero so we can divide out n
3x^2+3nx+n^2-29=0
solving for x we get
x = (-3n +- sqrt(372-3n^2))/6
so for x to be integer we need
372-3n^2>=0 and to be a perfect square, this leaves us with
3n^2<=372
n^2<=124
n<=11
trying out all possible values of n=1,2,3,...,11 we find that only
n=4,7,11 give positive perfect squares

Thus, the possible values for n are 4 7 and 11

Posted by Daniel on 2011-02-23 11:41:08