JOMO 6, Short 2

First, we must see ${ n }^{ 5 }-{ n }^{ 2 }={ n }^{ 2 }({ n }^{ 3 }-1)=25k$ for some $k$ . See that the only possibilities are ${ n }^{ 2 }$ is a multiple of 25 or ${ n }^{ 3 }-1$ is a multiple of 25. The other possibilities are ${ n }^{ 2 }$ is a multiple of $5$ with ${ n }^{ 3 }-1$ a multiple of $5$ too. But if ${ n }^{ 2 }$ is a multiple of $5$ then ${ n }^{ 3 }-1$ is not a multiple of $5$ ; and the same vice versa.

Now, lets check the first case ${ n }^{ 2 }$ is a multiple of 25. Then, ${ n }^{ 2 }=25x$ for some $x$ which implies $n=5\sqrt { x }$ . So here we obtain this sequence of solutions of $n$ considering $x$ as a square and $n<300$

$n=1, 5, 10, 15, 20, 25,..., 395$

a total of 60 solutions.

Second case, ${ n }^{ 3 }-1$ is a multiple of 25. See ${ n }^{ 3 }-1\equiv 0\quad (mod\quad 25)\Rightarrow { n }^{ 3 }\equiv 1\quad (mod\quad 25)\Leftrightarrow n\equiv 1\quad (mod\quad 25)$ . So, we have te solutions $n=1, 26, 51,..., 276$ a total of 11 (because we already have the answer $1$ ).

So $n$ has a total of $\boxed { 71 }$ solutions.

n^5-n^2 = n^2(n-1)(n^2 + n + 1) for i case : when 25/n^2 there are 59 no & for case 2: when 25/n-1 the possible values like 24, 49 & no of possible values are 12 but for case 3: when we replace n by unit integer there is no value which is divisible by 5 so in this case there is no value possible so no of values = 59 + 12=71