prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1: 
Try n = 0, we get:  $5 - 3 - 2 = 0$, which is divisible by 30. 
Try n = 1, we get: $5^{3} - 3^{3} - 2^{3} = 90$, which is also divisible by 30.
Step 2:
Assume it is true for n = k. 
So we are assuming the following equality is true: $5^{2k+1} - 3^{2k+1} - 2^{2k+1} = 30M$, for some integer M.
Step 3:
Now we look at the next case: n = k + 1.
$5^{2(k+1)+1} - 3^{2(k+1)+1} - 2^{2(k+1)+1}$
= $5^{2k+3} - 3^{2k+3} - 2^{2k+3}$
= $25\times5^{2k+1} - 9\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} + 4\times5^{2k+1} - 5\times3^{2k+1} - 4\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times[5^{2k+1} - 3^{2k+1} - 2^{2k+1}]$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times30M$.  (Assumed in step 2)
The last term is divisible by 30. But I cannot get a factor 30 out of the first two terms. I can show divisibility by 15 as follows:
= $7\times3\times5\times5^{2k} - 5\times3\times3^{2k} + 4\times30M$
= $7\times15\times5^{2k} - 15\times3^{2k} + 4\times15\times2M$
But how do I show divisibility by 30? 

Much the simplest method is to use arithmetic modulo 2,3 and 5.
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 1^{2n+1} - 1^{2n+1} - 0^{2n+1}\equiv 0 \pmod 2$$
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 2^{2n+1} - 0^{2n+1} - 2^{2n+1}\equiv 0 \pmod 3$$
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 0^{2n+1} - (-2)^{2n+1} - 2^{2n+1}\equiv 0\pmod 5$$
