finding all $(a,b)$

How to find all $(a,b)$ in $\mathbb{N}$ which $( 2a - 1 , 2b + 1 ) = 1$ and $a+b \mid 4ab + 1$

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Note that if $a+b$ divides $4ab+1$ then it also divides $4ab+2(a+b)+1$ which is $(2a+1)(2b+1)$. –  Gerry Myerson Oct 25 '12 at 12:26

If $4ab+1=k(a+b)$, then $(2a+1)(2b+1)=4ab+2(a+b)+1=(k+2)(a+b)$. This allows us to write $a+b=d_1d_2$ with $d_{1,2}\in\mathbb N$, $d_1|2a+1$ and $d_2|2b+1$. Similarly $(2a-1)(2b-1)=4ab-2(a+b)+1=(k-2)(a+b)$, hence $a+b=e_1e_2$ with $e_{1,2}\in\mathbb N$, $e_1|2a-1$ and $e_2|2b-1$.

As consecutive odd numbers are relatively prime, we have $(2a-1,2a+1)=(2b-1,2b+1)=1$. Therefore $(d_1,e_1)=(d_2,e_2)=1$. This implies $d_1=e_2$ and $d_2=e_1$. But then $(2a-1,2b+1)=1$, $e_1|2a-1$, $d_2|2b+1$ imply $d_2=e_1=1$, $d_1=e_2=a+b$. Thus $a+b|2a+1$ and $a+b|2b-1$, especially $a+b\le 2a+1\Rightarrow b\le a+1$ and $a+b\le 2b-1\Rightarrow a\le b-1$. Both together imply $b=a+1$. Therefore we are looking for $a\in\mathbb N$ such that

• $(2a-1,2a+3)=1$
• $2a+1|4a(a+1)+1=4a^2+4a+1=(2a+1)^2$

The first condition is always true because both numbers are odd and their difference is 4. The second condition is of course also always true. Therefore the general solution to the original problem is

$$b=a+1, a\in\mathbb N.$$

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$2a+1|4a^2+4a+1$ instead of $2a+1|4a^2+a+1$ –  Frank Oct 25 '12 at 15:08
Also observe that $2a+1|4a^2+4a+1$ implies $2a+1|2(2a+1)$ –  Frank Oct 25 '12 at 16:51
@MohammedAl-mubark oops. That slip kille dinfinitely many solutions. Thanks for the hint. –  Hagen von Eitzen Oct 25 '12 at 18:31

$\begin{eqnarray} {\bf Hint}\ \ \ \rm\color{#C00}{a\!+\!b} &|&\rm \color{#0A0}{4ab\!+\!1} \\ \rm\Rightarrow\ \color{#C00}{a\!+\!b} &|&\rm (2a\!+\!1)(2b\!+\!1) =\ \ 2(\color{#C00}{a\!+\!b}) + \color{#0A0}{4ab\!+\!1} \\ \Rightarrow\ \rm\color{#C00}{a\!+\!b} &|&\rm (2a\!+\!1)(2a\!-\!1) = 4a(\color{#C00}{a\!+\!b})\!-\!(\color{#0A0}{4ab\!+\!1}) \\ \Rightarrow\ \rm a\!+\!b &|&\rm (2a\!+\!1)(2a\!-\!1,2b\!+\!1) = 2a\!+\!1 \\ \Rightarrow\ \rm a\!+\!b &|&\rm (2b\!-\!1)(2a\!-\!1,2b\!+\!1) = 2b\!-\!1\ \ (analogous\ to\ above) \end{eqnarray}$

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