any better way to solve it?

Find all positive integers n such that n has exactly 6 positive

divisors 1 < d1 < d2 < d3 < d4 < n and 1 + n = 5(d1 + d2 + d3 + d4).

My solution is kind of too much effort. Any better way to do it?

1. there are two possible prime factorizations for n: n=p^5 or

n=p*q^2.

2. If n=p^5, 1+p^5=5(p+p^2+p^3+p^4).

1+p^5=(1+p)(p^4-p^3+p^2-p+1)=5(p^4+p^3+p^2+p). We have 5|(1+p) or

5|(p^4-p^3+p^2-p+1). p^4-p^3+p^2-p+1 = p(p-1)(p^2+1) + 1. if p=5m,

5m+1, 5m+2, 5m-2, 5|p(p-1)(p^2+1). So we can olny have p=5m-1 in both

cases. The smallest prime in form of 5m-1 is 19. if p >=19,

p+p^2+p^3+p^4 < p^2+p^2+p^3+p^4 < p^3+p^3+p^4 < p^4+p^4 = 2p^4. so

5(p+p^2+p^3+p^4) < 10p^4 < p^5+1. This means n cannot be p^5 for some

prime p.

3. So we must have n=p*q^2. Then we have 1+p*q^2=5p+5q+5pq+5q^2.

Rewrite the equation we have 1=5p+5q+5pq + q^2(5-p), so p>=7. Rewrite

the equation again as 1=5p+5q+5q^2 + pq(5-q), so q>=7.

4. From 1+p*q^2=5(p+q)(1+q), we know 20|1+p*q^2, the one's digits of p

 can only be 1 or 9 since the one's digit of q^2 cannot be 3 or 7.

This means p=10m+1 or p=10m-1.

5. One observation is that as q is increasing, q is approaching more

and more closer to 5. This needs to be proved and stated in a more

proper way.

6. If p=10m+1, lets assume q=10n+r, 1+p*q^2=1+(10m+1)*(10n+r)^2 =

1+r^2+10m*r^2 + 20(10m+1)(5*n^2+rn), since 20|1+p*q^2, so r=+/-3 and

m=2u+1. In other words, p=20u+11 and q=10n+/-3. q can be 7, 13, 17, 23

etc. If q=7, p=31 (good); if q=13, p=9.2 (invalid) and we can stop

here since the smallest possible p is 11.

7. If p=10m-1, lets assume q=10n+r, 1+p*q^2=1+(10m-1)*(10n+r)^2 =

1-r^2+10m*r^2 +20(10m-1)(5*n^2+rn), since 20|1+p*q^2, so r=+/-1 and

m=2u. In other words, p=20u-1 and q=10n+/-1. q can be 11, 19, 29, 31,

41 etc. If q=11, p=10.8 (invalid) and we can stop since the smallest

possible p is 19.

So the only qualified number n is 31*7*7.