Puzzle 778. A follow-up to Puzzle 774

Using Naohiro's conventions ...

Definitions:

1) s(n) = sigma(n). ( sigma is the sum of the divisors of n )

2) rad(n) = Largest square-free number dividing n.

Puzzle: Find a pair of numbers j & k, j≠k such that

s( j ) = s( k ) and

rad( j ) = rad( k )

Added twist:

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 solutions pairs

p1 and p2 (composed of n1 and m1, and n2 and m2) such that rad(n1) and rad(n2) are relatively prime,

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 solutions.

(Let's call these composites)

An example of this would be:

Solution pair for sigma = 210313800: 131576362 = 2 * 17 * 157^3 and 98731648 = 2^7 * 17^3 * 157

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) =

683686082027520000

So, this would be a level 2 composite solution.

Questions (Solutions must be non-trivial unless indicated otherwise):

Q1) What's the largest solution pair you can find?

Q2) Can you find a solution trio? If that's too difficult, can you find a trio where two members are "trivial"?

Q3) Can you a find a sigma with two distinct solution pairs.

Q4) Can you find a level 3 (or higher) composite solution?