Belarusian Mathematical Olympiad

Answer: b) there are an infinite number of nice numbers.

a) Let $N$ be a nice number, i.e. $N=d_1^4+d_2^4+d_3^4+d_4^4+d_5^4$, where $d_i$, $i=1,2,3,4,5$, are the distinct divisors of $N$. If some divisor of $N$ is divisible by $5$, then $N$ is divisible by $5$. So we suppose that $d_1,d_2,d_3,d_4,d_5$ are not divisible by $5$. Then their fourth powers are congruent to $1$ modulo $5$. Indeed, if $M=5k\pm 1$ ($k\in \mathbb{N}$), then $M^2=25k^2\pm 10k+1=5m+1$ ($m\in \mathbb{N}$) and $M^4=(5m+1{)}^2=25m^2+10m+1$, i.e. $M^4$ is congruent to $1$ modulo $5$; if $M=5k\pm 2$ ($k\in \mathbb{N}$), then $M^2=25k^2\pm 20k+4=5m+4$ ($m\in \mathbb{N}$) and $M^4=(5m+4{)}^2=25m^2+40m+16$, i.e. $M^4$ is congruent to $1$ modulo $5$. Therefore, the fourth power of any divisor $d_i$ is congruent to $1$ modulo $5$, i.e. $d_i^4=5k_i+1$, and it follows that $N$ is divisible by $5$.

b) Let there exist a nice number $N$, i.e. $N=d_1^4+d_2^4+d_3^4+d_4^4+d_5^4$, where $d_i$ are some positive integers and $p>1$. If $d_1,d_2,d_3,d_4,d_5$ are distinct divisors of $N$, consider the number $N(p)=N{p}^4$, where $p$ is any positive integer greater than $1$. It is obvious that $d_{1}p,d_{2}p,d_{3}p,d_{4}p,d_{5}p$ are distinct divisors of $N(p)$, and $$\begin{array}{rl}N(p)& =N{p}^4=[N=d_1^4+d_2^4+d_3^4+d_4^4+d_5^4]=(d_1^4+d_2^4+d_3^4+d_4^4+d_5^4)p^4=\\ & =(d_{1}p{)}^4+(d_{2}p{)}^4+(d_{3}p{)}^4+(d_{4}p{)}^4+(d_{5}p{)}^4.\end{array}$$ So there are infinitely many nice numbers $N(p)$ if there exists at least one nice number. It remains to note that the number $N=1^4+2^4+3^4+6^4+34^4$ is nice. Indeed, it is evident that $N$ is divisible by $2$. Since the numbers $1^4,2^4$, and $34^4$ are congruent $1$ modulo $3$, the number $N$ is divisible by $3$. Since $N$ is even and divisible by $3$, we see that $N$ is divisible by $6$. Besides, $$N=1^4+2^4+3^4(1^4+2^4)+34^4=(1^4+2^4)(1^4+3^4)+34^4=17\cdot (1^4+3^4)+34^4,$$ whence we see that $N$ is divisible by $34$. Thus, $N$ is nice.