A minute question

Number Theory Level 2

The expression $({ x }^{ mn }+1)$ is divisible by $(x+1)$ only of-

m is odd Both n and m are odd n is odd Can't say

2 solutions

Samarpit Swain
Aug 31, 2014

Let f(x)= x^(mn) +1, and g(x)=(x+1)

Now, for g(x) to be a FACTOR of f(x)....

f(-1)=0 [ By Factor theorem ].

Therefore, (-1)^(mn)+1=0 ,i.e, (-1)^(mn)= (-1)

For the above equation to hold true, mn has to be 'odd'.

We know that product of 2 numbers will be odd only when both the 2 numbers are odd.

[ As all other combinations will have an 'even' factor]

Hence, we finally conclude that both 'm' and 'n' are odd :)

Shivam S. Gour
Dec 9, 2014

Any expression (1 - x^y) , y is a natural number can be expressed in the form

            (1 - x^y) = (1 - x) <(1 + x)^(y-1)>

where in some notation <(1-a)^b> means considering only variables wihout their coefficients

So, in the given que. we replace

                     x ==> by (-x)

so we have to show that

               [1 + (-x)^mn] is divisible by  (1 - x)

But, to obtain the desired form

                   m*n must be  odd

But product of two number can only be odd iff both the numbers are odd Therefore,

                    m and n are both odd