Quote:
Originally Posted by xilman
Solve the equation 10^{25}x = 1 (mod 10^{50}) in integers. The solution is the integer you want.

Be careful! There exist no solution to this equation. Since 10
^{25} is even, a multiple of it is always even, and on the right hand side, 1 (mod 10
^{50}) is always odd. A solution is impossible to exist!
Quote:
Originally Posted by xilman
Substitute x = y + 1/y in the octic and see what you get ...

No hopes for degree 4. Substituting x = y + (1/y) in x
^{8}, so it gives up
which is clearly being at degree 8.
Other terms will have their appropriate degrees. So, when substituted, the whole algebraic polynomial will be of degree 8 only.
And the linear polynomial becomes more cumbersome, in this form, with
10
^{25}(y+(1/y))  (10
^{50}+1)