Find all positive integers $n$ such that $n$ has exactly $6$ positive divisors $1<d_{1}<d_{2}<d_{3}<d_{4}<n$ and $1+n=5(d_{1}+d_{2}+d_{3}+d_{4})$.
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Tags: calculus, integration, modular arithmetic, Divisibility Theory
Rust
16.10.2007 13:56
Obviosly $ n=p^5$ is not solution, therefore $ n=pq^2=5(p+pq+q+q^2)-1$ or $ p(q^2-5q-5)=5q(q+1)-1$, therefore $ p>5,q>5$. If $ q=7$ we get $ p=31$ (solution) if $ p=7$ we get contradition $ q(q-20)=17$. Because for $ q\ge 11$ we get $ p\le 7$, solution $ n=31*7^2$ is unique.
minimario
24.02.2015 06:10
We know that $n=p^5$ or $pq^2$. If $n=p^5$, then $1+p^5=5(p+p^2+p^3+p^4) \Rightarrow p^5-5p^4-5p^3-5p^2-5p+1=0$, which obviously has no prime roots $p$. (RRT does the trick).
If $n=pq^2$, we have \[pq^2+1=5p+5q+5pq+5q^2 \Rightarrow p = \frac{5q^2+5q-1}{q^2-5q-5} = 5+\frac{30q+24}{q^2-5q-5}\]
Note that if $q=2, 3, 5$, $p$ is not integral. Otherwise, $p$ is positive. Let $f(x)=\frac{30q+24}{q^2-5q-5}$. Note that $f$ is decreasing from $5$ to infinity, and that $f(7)=26$ and $f(11)\approx 5.8$. If there is a solution, then $k \le 5$ Additionally, note that if $\frac{30q+24}{q^2-5q-5} = k$ , then $kq^2-5kq-5=30q+24 \Rightarrow kq^2-(30+5k)q-(5k+24)=0$. If there are integral solution, $\Delta = 25k^2+300k+900+20k^2+96=3(15k^2+68k+300)$. If this is a perfect square, $15k^2+68k+300 \equiv 0 \pmod 3 \Rightarrow k \equiv 0 \pmod 3 \Rightarrow k = 3$. However, obviously there is no solution when $k=3$, so the only solution is $(p, q) = (31, 7)$, or $n=1519$.