- **Math**
(*https://www.mersenneforum.org/forumdisplay.php?f=8*)

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

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. |

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

[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: |

[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: |

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=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: |

[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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