20050411, 13:36  #1 
Apr 2005
2·19 Posts 
Possible solutions to an equation:
What would the possible solution(s) to the following equation be:
(pq + r)^r = 2^(p + r^2) where r is an even number. 
20050411, 13:53  #2 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
q=(2^((p+r^2)/r)r)/p according to Maple, and easy to check.
Alex 
20050411, 14:44  #3  
Nov 2003
2^{2}×5×373 Posts 
Quote:
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^(mh), 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. 

20050411, 15:13  #4  
Apr 2005
2·19 Posts 
Quote:
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? 

20050413, 03:00  #5 
"William"
May 2003
New Haven
2^{3}·5·59 Posts 
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^{mh} If we let z=mh, we have a solution for any values of z, k, q, and h with k=z2^{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)/(z2^{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 
20050413, 19:23  #6  
Apr 2005
2·19 Posts 
Quote:


20050414, 05:19  #7  
"William"
May 2003
New Haven
2^{3}·5·59 Posts 
Quote:
x=2 (it must be a factor of 2^{h}h+1=2) k=1 (k and q must be factors of 2^{x}1=3) y=1 z=2 q=3 m=2 r=1 p=1 For the tiny solution p=1, q=3, r=1, and the equation become 4=4. Now for a reverse challenge  how many solutions can you find between this tiny solution and my example? 

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