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Old 2005-04-11, 14:44   #3
R.D. Silverman
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Nov 2003

22·5·373 Posts

Originally Posted by akruppa
q=(2^((p+r^2)/r)-r)/p according to Maple, and easy to check.

I would assume that this equation is Diophantine.

It is clear that pq+r must be a power of 2, say 2^m, whence mr = p+r^2
and thus m = (p+r^2)/r and thus p must be divisible by r.

Thus p = kr giving us m = k+r and qkr + r = 2^m, thus r must be
a power of two as well, (say) 2^h But this then yields qk 2^h + 2^h = 2^m
or qk+1 = 2^(m-h), Thus qk+1 must be a power of 2.
Putting this together, we get:

(2^h)^(2^h) (qk+1)^(2^h) = 2^(k2^h + 2^2h) I would expect solutions to be few and far between; I will look more closely later.
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