Question

show that x+a is a factor of x power n +a power n

Answer 1

∴ When ’n’ is odd x+yx+y is a factor of xn+ynxn+yn.

Using Mathematical Induction

If n=1n=1, then, xn+yn = x1+y1 = x+yxn+yn = x1+y1 = x+y, which is divisible by x+yx+y.

Let us assume that xn+ynxn+yn is divisible by x+yx+y for some odd number 2k−12k−1and assume that the quotient is pp.

From this, we get,
x2k−1+y2k−1 = p(x+y)  ⇒  y2k−1 = p(x+y) − x2k−1x2k−1+y2k−1 = p(x+y)  ⇒  y2k−1 = p(x+y) − x2k−1

Now, we need to prove that this is also true for the next odd number i.e. for n=2k+1n=2k+1.

Consider x2k+1+y2k+1x2k+1+y2k+1

= x2⋅x2k−1+y2⋅y2k−1= x2⋅x2k−1+y2⋅y2k−1

= x2⋅x2k−1+y2(p(x+y) − x2k−1)= x2⋅x2k−1+y2(p(x+y) − x2k−1)

= x2⋅x2k−1+py2(x+y) −y2⋅x2k−1= x2⋅x2k−1+py2(x+y) −y2⋅x2k−1

= (x2−y2)x2k−1+py2(x+y)= (x2−y2)x2k−1+py2(x+y)

= (x+y)(x−y)x2k−1+py2(x+y)= (x+y)(x−y)x2k−1+py2(x+y)

= (x+y)[(x−y)x2k−1+py2]= (x+y)[(x−y)x2k−1+py2]

= (x+y)q= (x+y)q

So, it is proved that x2k+1+y2k+1x2k+1+y2k+1 is divisible by x+y.x+y.

∴∴ According to the principle of Mathematical Induction, x+yx+y is a factor of xn+ynxn+yn for every odd integer ‘n’.

Using the Remainder theorem

Let f(x) = xn+ynf(x) = xn+yn.

The remainder when f(x) is divided by x+yx+y is f(−y)f(−y).

f(−y) = (−y)n+ynf(−y) = (−y)n+yn

For perfect division the remainder must be zero. So, here the value of f(−y)f(−y)must be zero, which is possible only when ’n’ is odd.

For odd ‘n’,
f(−y) = (−y)n+yn = −yn+yn=0