Prove $n^2(n^4-1)$ is divisible by 60 using Mathematical Induction. Base step:
p(2)=4 * 15= 60
Inductive Hypothesis:
Assuming p(k) = $k^2(k^4-1)$ = 60q
Induction:
p(k+1)= $(k+1)^2[(k+1)^4-1]$
      = $(k+1)^2[(k+1)^2 + 1][(k+1)^2 - 1]$
      = $(k+1)^2(k^2+2k+2)(k^2+2k+1- 1)$
      = $(k+1)^2(k^2+2k+2) k (k+2)$
      = $k(k+1)(k+2)(k^2+2k+2)(k+1)$
Now $k(k+1)(k+2)$ is the product of 3 consecutive natural numbers. Their product HAS to be a multiple of 2 and 3 and hence also a multiple of 6. So all that I have to do is prove that $(k^2+2k+2)(k+1)$ is a multiple of 10. But I can't seem to do so. Also through my induction, I have not at all made use of my hypothesis! How do I go about solving this question through the Principle of Mathematical Induction?

If $f(n)=n^2(n^4-1)=n^6-n^2$ is divisible by $60$ for $n=k$
$$f(k+1)-f(k)=(k+1)^6-(k+1)^2-\{k^6-k^2\}$$
$$=6k^5+15k^4+20k^3+15k^2+6k-2k$$
which is $\displaystyle6(k^5-k)+15k^4+20k^3+15k^2+10k$ divisible by $5$ as $k^5-k$
Again, $\displaystyle6k^5+15k^4+20k^3+15k^2+4k\equiv2(k^3-k)\pmod3\equiv0$
finally, $\displaystyle6k^5+15k^4+20k^3+15k^2+4k=2k^5-k^4-k^2\pmod4\equiv2k^4(k-1)+k^4-k^2$
$\displaystyle\equiv2k^3 \underbrace{k\cdot(k-1)}_{\text{ even}}+k^2(k-1)(k+1)\pmod4$
If $k$ is even, $\displaystyle4|k^2$ else $2|(k\pm1)\implies 2^2|(k^2-1)$
So, $\displaystyle f(k+1)-f(k)$ will be divisible by lcm$(4,3,5)=60$
