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Patron · **Joined:** Tue, 1 Nov 2011 00:12:51 UTC **Posts:** 6

$a=\frac{62k+1}{k-1}$

a, k are natural numbers
I must find all a

Shadow · **Posted:** Tue, 1 Nov 2011 20:48:24 UTC

Patron wrote:

$a=\frac{62k+1}{k-1}$

a, k are natural numbers
I must find all a

multiply both sides by k-1 (justify why this is OK), solve linear equation.

Denis · **Posted:** Tue, 1 Nov 2011 21:30:07 UTC

solutions infinite (except a<>62).

Patron · **Joined:** Tue, 1 Nov 2011 00:12:51 UTC **Posts:** 6

I knows results. a=63, 65, 69, 71 ,83, 125, but my system is very complicated.

Shadow · **Posted:** Tue, 1 Nov 2011 22:43:21 UTC

Denis wrote:

solutions infinite (except a<>62).

What? For fixed

and $k\ne 1$ (for which the right side is not defined, so this is not well-defined) we can see that $62k+1=a(k-1)\iff k(62-a)+1+a=0$ , which has no solution for a=62

as you note, but otherwise, it's easy to see there is a unique solutions of $$k=-{a+1\over 62-a}$ , which is uniquely determined by

. . .

I mean in general this should be obvious because the given function is a Möbius transform, which is an analytic isomorphism of the Riemann sphere, so if you consider the punctured sphere, you need to only punch out the preimage of infinity, in this case a=62

and you're fine, and you still get a 1-1 function descending from the original by restriction.

Soroban · **Posted:** Wed, 2 Nov 2011 02:40:25 UTC

Hello, Patron!

Quote:

$a\:=\:\dfrac{62k+1}{k-1} \quad a, k\text{ are natural numbers. \; Find all }a.$

We have: . $a \:=\:62 + \dfrac{63}{k-1}$

Since

is a natural number, k-1

must be a factor of 63.

And we have six solutions: . $\begin{array}{cc} k & a \\ \hline 64 & 63 \\ 22 & 65 \\ 10 & 69 \\ 8 & 71 \\ 4 & 83 \\ 2 & 125 \end{array}$

Denis · **Posted:** Wed, 2 Nov 2011 04:36:54 UTC

Ok ok .... missed the word "natural" ... :cry:

Shadow · **Posted:** Wed, 2 Nov 2011 04:37:37 UTC

Denis wrote:

Ok ok .... missed the word "natural" ... :cry:

As did I, but I'm still not sure how you got infinitely many solutions?

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