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Old 2005-04-11, 15:13   #4
Apr 2005

2·19 Posts

Originally Posted by R.D. Silverman
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.
Thats true, this is a Diophantine equation.
I am currently investigationg and creating diophantine equations.
And you are quite right in your working, very clear indeed.

What's your opinion on Diophantine equations, do they interest you?
Vijay is offline   Reply With Quote