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 2005-04-11, 13:36 #1 Vijay   Apr 2005 3810 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.
 2005-04-11, 13:53 #2 akruppa     "Nancy" Aug 2002 Alexandria 25×7×11 Posts q=(2^((p+r^2)/r)-r)/p according to Maple, and easy to check. Alex
2005-04-11, 14:44   #3
R.D. Silverman

Nov 2003

26·113 Posts

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

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?

 2005-04-13, 03:00 #5 wblipp     "William" May 2003 New Haven 234010 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+2h and (kq+1)=2m-h If we let z=m-h, we have a solution for any values of z, k, q, and h with k=z-2h+h and kq=2z-1 Eliminating k, we see that we have a solution for any choice of z, q, and h with q=(2z-1)/(z-2h+h) z cannot be a prime number because the denominator is smaller than z and all factors of 2p-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 2x-1. So to generate a solution: 1. Pick a value for h. 2. Pick x to be a factor of 2h-h+1 3. Pick the denominator to be a factor of 2x-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. 2h-h+1=3, so pick x=3 3. 2x-1=7, so pick the denominator=7 4. y = 3 5. z = xy = 9 6. q = (2z-1)/denominator = 73 7. m = z+h = 11 8. k = denominator = 7 9. r = 2h= 4 10. p = kr = 28
2005-04-13, 19:23   #6
Vijay

Apr 2005

2×19 Posts

Quote:
 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+2h and (kq+1)=2m-h If we let z=m-h, we have a solution for any values of z, k, q, and h with k=z-2h+h and kq=2z-1 Eliminating k, we see that we have a solution for any choice of z, q, and h with q=(2z-1)/(z-2h+h) z cannot be a prime number because the denominator is smaller than z and all factors of 2p-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 2x-1. So to generate a solution: 1. Pick a value for h. 2. Pick x to be a factor of 2h-h+1 3. Pick the denominator to be a factor of 2x-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. 2h-h+1=3, so pick x=3 3. 2x-1=7, so pick the denominator=7 4. y = 3 5. z = xy = 9 6. q = (2z-1)/denominator = 73 7. m = z+h = 11 8. k = denominator = 7 9. r = 2h= 4 10. p = kr = 28
Quite nicely worked out, but there does exist a smaller 'friendly' solution. Can anybody find the small solution?

2005-04-14, 05:19   #7
wblipp

"William"
May 2003
New Haven

92416 Posts

Quote:
 Originally Posted by Vijay Quite nicely worked out, but there does exist a smaller 'friendly' solution. Can anybody find the small solution?
There should be several smaller solutions. My example was intended to be small but informative. Tiny solutions make poor examples because everything tends to fold in on itself. I would expect the smallest solution to come from setting h=0. That leads to the choices

x=2 (it must be a factor of 2h-h+1=2)
k=1 (k and q must be factors of 2x-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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