Prove 5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33.

Prove 5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33.

I wonder if there are other ways that can prove this.

My way of proving this:

Let P(n) be "5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33." for all positive integers n.

When n = 1, 5^3+11^3+17^3 = 33*193

Thus P(1) is true.

Assume P(k) is true for some positive integers k.

i.e. 5^(2k+1)+11^(2k+1)+17^(2k+1) = 33N, where N is an integer.

When n = k+1,

5^[2(k+1)+1] + 11^[2(k+1)+1] + 17^[2(k+1)+1]

=5^(2k+3) + 11^(2k+3) + 17^(2k+3)

=25[5^(2k+1)] + 121[11^(2k+1)] + 289[17^(2k+1)]

=25[5^(2k+1)+11^(2k+1)+17^(2k+1)] + 96[11^(2k+1)] + 264[17^(2k+1)]

=25(33N) + 99[11^(2k+1)] - 3[11^(2k+1)] + 264*17^(2k+1)

=33[25N + 3*11^(2k+1) - 11^(2k) + 8*17^(2k+1)]

Since N and k are integers,

25N + 3*11^(2k+1) - 11^(2k) + 8*17^(2k+1) is an integer.

Therefore, 25N + 3*11^(2k+1) - 11^(2k) + 8*17^(2k+1) is divisible by 33.

Thus P(k+1) is true.

By M.I., P(n) is true for all positive integers n.

Re: Prove 5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33.

I think such type of questions are best dealt with by Mathematical Induction.

Re: Prove 5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33.

Another way

The given expression is divisible by 11 using the fact that 5 = -6 (mod 11) and 17 = 6 (mod 11)

Similarly, we can use congruence mod 3 to show that the expression is divisible by 3

Therefore it is divisible by 33

Re: Prove 5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33.

Thanks, Idea!

Are there no other ways of proving this?

Re: Prove 5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33.

How many do you need? Proof by induction, which is what you did, is what I would have used but I agree that a proof by congruence is simpler.

Re: Prove 5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33.

I just want to know if there are other ways to prove it. Do you prove it by induction in the same way as I did? If not, I would like to know how you do it. Thanks!