[Vijay]

Originally Posted by wblipp

One small solution is p=28, q=73, r=4.

It's easy to generate an infinite number of solutions from Mersenne factors, although they grow rather quickly.

When Robert Silverman left off, we have a solution for any choices of m. k, q, and h such that

m=k+2^h and (kq+1)=2^m-h

If we let z=m-h, we have a solution for any values of z, k, q, and h with

k=z-2^h+h and kq=2^z-1

Eliminating k, we see that we have a solution for any choice of z, q, and h with

q=(2^z-1)/(z-2^h+h)

z cannot be a prime number because the denominator is smaller than z and all factors of 2^p-1 are of the form 2ap+1, and therefor larger than p.

So let z=xy and pick the denominator to be a factor of 2^x-1.

So to generate a solution:
1. Pick a value for h.
2. Pick x to be a factor of 2^h-h+1
3. Pick the denominator to be a factor of 2^x-1
4. Solve for y from the denominator choice
5. z=x*y
6. Work backwards to p, q, and r.

For example
1. h=2
2. 2^h-h+1=3, so pick x=3
3. 2^x-1=7, so pick the denominator=7
4. y = 3
5. z = xy = 9
6. q = (2^z-1)/denominator = 73
7. m = z+h = 11
8. k = denominator = 7
9. r = 2^h= 4
10. p = kr = 28

Quite nicely worked out, but there does exist a smaller 'friendly' solution. Can anybody find the small solution?