integral solutions

???

If someone has the patience to find solutions for x=6,7,8, then there is a simple solution possible.

Say we have a solution for n=k,

\frac{1}{x_1^2} + \frac{1}{x_2^2}+...+\frac{1}{x_k^2} =1

Then \frac{1}{4x_1^2} + \frac{1}{4x_2^2}+...+\frac{1}{4x_k^2} =\frac{1}{4}

and so

\frac{1}{4x_1^2} + \frac{1}{4x_2^2}+...+\frac{1}{4x_k^2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} =1

i.e. we have the solution (2x_1, 2x_2,...,2x_k,2,2,2) for n = k+3.

Turns out, the solutions for the three cases are easy to set up (in other words i was plain lazy)

n=6:

\left(\frac{1}{4} + \frac{1}{9} \right)+ \left(\frac{1}{4} + \frac{1}{9} \right)+\left(\frac{1}{4} + \frac{1}{36} \right)=1

n=7:

\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4} = 1

n=8:

\left(\frac{1}{4} + \frac{1}{9} \right)+ \left(\frac{1}{4} + \frac{1}{9} \right)+\left(\frac{1}{9} + \frac{1}{36} \right)+\left(\frac{1}{9} + \frac{1}{36} \right) = 1

thank u sir!

We have an alternative soln if the multiplication by 1/4 does not hit.

Recollect the fact that a square can be dividen in n non-overlapping squares for all n≥6.

Take the unit squares.

If we say 1x_i=a_ii.

We see each a_i is either of the form 12^k, where k are non negative integers. Basically these a_is arethe lengths of the squares into which the unit square has been divided.

The fact that the above qn has integral solutions follow from the fact that since these squares are non-overlapping, their sum of areas must be equal to the area of the square which was partitioned, i.e 1.

Heck ya! The connection with that prob though nagging at me did not occur this clearly to me.

The problem referred to is this: http://targetiit.com/iit-jee-forum/posts/dividing-squares-12381.html

yeah its opening now ( i had blocked tiit on my pc due to addiction [4] )