1) s(n) = sigma(n). (
sigma is the sum of the divisors of n )
2) rad(n) = Largest square-free number
Puzzle: Find a pair of
numbers j & k, j≠k such that
s( j ) = s( k ) and
rad( j ) = rad( k )
Because of the property
where if n != m where n and m are relatively prime to x,
if sigma(n) = sigma(m), then sigma(n*x) = sigma(m*x), we
must limit ourselves to non-trivial answers.
If p divides both n and m,
then p cannot have the same exponent in the
factorization of m and n. (The same relation does not
hold for sigma(n) - n)
The smallest solution to this problem is sigma = 4800:
2058 = 2 * 3 * 7^3 and 1512 = 2^3 * 3^3 * 7
(This on the other hand,
would be a trivial solution:
sigma = 4800*(5+1): 2058*5 = 2 * 3 * 5 *
7^3 and 1512*5 = 2^3 * 3^3 * 5 * 7
That's because we took an
existing solution non-trivial pair and multiplied each
member by 5)
We can extend this idea of
triviality to this problem itself. If we have two
p1 and p2 (composed of n1
and m1, and n2 and m2) such that rad(n1) and rad(n2) are
then n1 * m1 and n2 * m2
would form a solution pair. Similarly, so would n2 * m1
and n1 * m2.
Both would be considered
trivial because they were piggybacking on other
(Let's call these
An example of this would
Solution pair for sigma = 210313800:
131576362 = 2 * 17 * 157^3 and 98731648 = 2^7 * 17^3 *
Solution pair for sigma = 3250790400: 2196937295 =
5 * 7^3 * 31^3 * 43 and 2156627375 =
5^3 * 7 * 31 * 43^3
Clearly, the rads for the
two pairs have no factors in common so we have these
"trivial" solutions below.
sigma(131576362 * 2196937295)
= sigma(98731648 * 2156627375)
sigma(131576362 * 2156627375)
= sigma(98731648 * 2196937295)
So, this would be a level
2 composite solution.
Questions (Solutions must be non-trivial unless
What's the largest solution pair you can find?
you find a solution trio? If that's too difficult, can
you find a trio where two members are "trivial"?
you a find a sigma with two distinct solution pairs.
you find a level 3 (or higher) composite solution?