Problems & Puzzles: Puzzles

Puzzle 370. The first 13 numbers

Muneer Jebreel Karama sent the following puzzle:

He sent one solution with (p,q,r,s) = (5,3,2 11)


Find more solutions.


Jan Van Delden wrote (June 2007):

I'm rather interested in the motivation underlying this problem, apart from the fact that one solution is easy to find. And apparently another quite hard. (Although I didn't use a number cruncher to be frank; I tested for p,q <50, very small for this site's standards I admit).
It seems to me that the restrictions don't allow for a lot of solutions. If we allow to loosen the restrictions it still will be difficult to find large solutions.
With: S_p[n]=Sum(i^p,i=1..n)
The problem states:
I)    S_p[13]=s^r S_q[13]
II)   S_p[13]=k^r
The first restriction will be hard enough as it is.
First point:
For the second restriction:
S_p[n] = k^2 seems to give a solution with p=3 for all n.
This is no surprise since S_3[n]=S_1[n]^2; Nichomachus's theorem.
Second point:
S_p[13] will be of the form 7^k * 13^l * remaining term, k and l>=1.
[If n prime then S_p[n] is congruent to sum(i,i=1..n) mod p, which gives S_p[13] congruent to 7*13 mod p, so at least 0 mod 7 and 0 mod 13]
Testing for several p shows the remaining term is not very likely to have the same factors, where the biggest one grows rather fast. Furthermore the the multiplicity of the factors 7 and 13 don't appear to have a 'regular' pattern (I missed it).
So we might be 'lucky' to find a second solution with r=2, let alone a different r>2 prime.
If we alter the problem in the form:
S_p[s] = r S_q[s] with p,q,r,s prime
It's not that hard to find small solutions for certain combinations of p,q;
For instance:
It will be hard enough to find large ones.
Even finding large solutions to t S_p[s] = r S_q[s] with p,q,r,s,t prime will require some crunching.
Then again, I'm not a number theorist, but as posed it looks a bit too much 'Wieferich' to me.



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