Plzzz prove this... divisibility !!!!!!!!?

Probably the easiest way to get that result is modular arithmetic.

Modulo 22, 9 and 13 are opposites:

13 == -9 mod 22

because 13 + 9 = 22

So raising both to any odd power,

13^(2n+1) == (-9)^(2n+1) == (-1)^(2n+1) • 9^(2n+1) == -[9^(2n+1)]

thus,

13^(2n+1) + 9^(2n+1) == 0 mod 22

QED

EDIT:

This same method will prove Madhukar's generalization of this proposition.

@ oregfiu: Nice proof! But the starting condition is even easier with n=0 !

oregfiu's method will also prove Madhukar's generalized proposition.

Let 2n + 1 = m

Obviously, m is an odd number

=> 13^m + 9^m

= (13 + 9) * [(13)^(m-1) - (13)^(m-2) * (9) + (13)^(m-3) * (9)^2 - ... + (9)^(m-1)]

As this has a factor 22, the number is divisible by 22.

Edit:

In general, one can say that

a^m + b^m is divisible by (a + b) if m is odd.

Proof by mathematical induction:

1st step:

For n=1

13^(2*1 + 1) + 9^(2*1 + 1) = 2926 = 22 * 133

2nd step:

Suppose that

13^(2k + 1) + 9^(2k + 1) = 22 N

where N is natural number.

Then

13^(2(k+1) + 1) + 9^(2(k+1) + 1) =

= 13^2 * 13^(2k + 1) + 9^2 * 9^(2k + 1) =

= 169 * 13^(2k + 1) + 81 * 9^(2k + 1) =

= (81 + 88) * 13^(2k + 1) + 81 * 9^(2k + 1) =

= 81 * 13^(2k + 1) + 81 * 9^(2k + 1) + 88 * 13^(2k + 1) =

= 81 (13^(2k + 1) + 9^(2k + 1)) + 88 * 13^(2k + 1) =

= 81 * 22 N + 88 * 13^(2k + 1) =

= 22 (81 N + 4 * 13^(2k + 1))

Q.E.D.

-