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2005-04-11, 15:13   #4
Vijay

Apr 2005

2·19 Posts

Quote:
 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?