(a) If one of $a,b,c,d,e$ is divisible by $5$, then also $n$ (as a multiple of $a,b,c,d,e$) is divisible by $5$. If none of $a,b,c,d,e$ is divisible by $5$, then $a^4,b^4,c^4,d^4,e^4$ are all $\equiv1\bmod5$ by Fermat. Then $n=a^4+b^4+c^4+d^4+e^4\equiv1+1+1+1+1\equiv0\bmod5$. (b) Let $a'<b'<c'<d'<e'$ be five arbitrary positive integers, and let $S=(a')^4+(b')^4+(c')^4+(d')^4+(e')^4$ and let $P=a'b'c'd'e'$. Then $S\cdot P^4=(a'P)^4+(b'P)^4+(c'P)^4+(d'P)^4+(e'P)^4$. As $a'P,b'P,c'P,d'P,e'P$ each divide $S\cdot P^4$, we see that $n=S\cdot P^4$ is marvelous. Hence there are infinitely many marvelous numbers.

Can someone check my solution for $(b)$? Let $n=(5^k)^4+(5^{k+1})^4+(5^{k+2})^4+(5^{k+3})^4+(5^{k+4})^4$. It is obvious that $5^{4k}|n$. So if $4k \geq k+4$ then $5^{k+1}, 5^{k+2}, 5^{k+3}, 5^{k+4}$ are also divisors of $n$. For any integer value of $k$ such $k>1$ we can get infinitely marvelous numbers.

BarishNamazov wrote: Can someone check my solution for $(b)$? Let $n=(5^k)^4+(5^{k+1})^4+(5^{k+2})^4+(5^{k+3})^4+(5^{k+4})^4$. It is obvious that $5^{4k}|n$. So if $4k \geq k+4$ then $5^{k+1}, 5^{k+2}, 5^{k+3}, 5^{k+4}$ are also divisors of $n$. For any integer value of $k$ such $k>1$ we can get infinitely marvelous numbers. Bump...

BarishNamazov wrote: BarishNamazov wrote: Can someone check my solution for $(b)$? Let $n=(5^k)^4+(5^{k+1})^4+(5^{k+2})^4+(5^{k+3})^4+(5^{k+4})^4$. It is obvious that $5^{4k}|n$. So if $4k \geq k+4$ then $5^{k+1}, 5^{k+2}, 5^{k+3}, 5^{k+4}$ are also divisors of $n$. For any integer value of $k$ such $k>1$ we can get infinitely marvelous numbers. Bump... Yes that is correct.