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-   -   Possible solutions to an equation: (https://www.mersenneforum.org/showthread.php?t=3989)

Vijay 2005-04-11 13:36

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.

akruppa 2005-04-11 13:53

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

Alex

R.D. Silverman 2005-04-11 14:44

[QUOTE=akruppa]q=(2^((p+r^2)/r)-r)/p according to Maple, and easy to check.

Alex[/QUOTE]

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. :question:

Vijay 2005-04-11 15:13

[QUOTE=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. :question:[/QUOTE]

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? :cool:

wblipp 2005-04-13 03:00

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[sup]h[/sup] and (kq+1)=2[sup]m-h[/sup]

If we let z=m-h, we have a solution for any values of z, k, q, and h with

k=z-2[sup]h[/sup]+h and kq=2[sup]z[/sup]-1

Eliminating k, we see that we have a solution for any choice of z, q, and h with

q=(2[sup]z[/sup]-1)/(z-2[sup]h[/sup]+h)

z cannot be a prime number because the denominator is smaller than z and all factors of 2[sup]p[/sup]-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[sup]x[/sup]-1.

So to generate a solution:
1. Pick a value for h.
2. Pick x to be a factor of 2[sup]h[/sup]-h+1
3. Pick the denominator to be a factor of 2[sup]x[/sup]-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[sup]h[/sup]-h+1=3, so pick x=3
3. 2[sup]x[/sup]-1=7, so pick the denominator=7
4. y = 3
5. z = xy = 9
6. q = (2[sup]z[/sup]-1)/denominator = 73
7. m = z+h = 11
8. k = denominator = 7
9. r = 2[sup]h[/sup]= 4
10. p = kr = 28

Vijay 2005-04-13 19:23

[QUOTE=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+2[sup]h[/sup] and (kq+1)=2[sup]m-h[/sup]

If we let z=m-h, we have a solution for any values of z, k, q, and h with

k=z-2[sup]h[/sup]+h and kq=2[sup]z[/sup]-1

Eliminating k, we see that we have a solution for any choice of z, q, and h with

q=(2[sup]z[/sup]-1)/(z-2[sup]h[/sup]+h)

z cannot be a prime number because the denominator is smaller than z and all factors of 2[sup]p[/sup]-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[sup]x[/sup]-1.

So to generate a solution:
1. Pick a value for h.
2. Pick x to be a factor of 2[sup]h[/sup]-h+1
3. Pick the denominator to be a factor of 2[sup]x[/sup]-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[sup]h[/sup]-h+1=3, so pick x=3
3. 2[sup]x[/sup]-1=7, so pick the denominator=7
4. y = 3
5. z = xy = 9
6. q = (2[sup]z[/sup]-1)/denominator = 73
7. m = z+h = 11
8. k = denominator = 7
9. r = 2[sup]h[/sup]= 4
10. p = kr = 28[/QUOTE]

Quite nicely worked out, but there does exist a smaller 'friendly' solution. Can anybody find the small solution? :smile:

wblipp 2005-04-14 05:19

[QUOTE=Vijay]Quite nicely worked out, but there does exist a smaller 'friendly' solution. Can anybody find the small solution?[/QUOTE]

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 2[sup]h[/sup]-h+1=2)
k=1 (k and q must be factors of 2[sup]x[/sup]-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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