a) the equation is symmetric in $a$ and $b$. $(2,3)$ is a solution with $k=2$. by vieta-jumping we get for the 2 zeroes $a_{1}$ and $a_{2}$ of $a^{2}+b^{2}-1=kab$ the equations $a_{1}+a_{2}=kb$ and $a_{1}a_{2}=b^{2}-1$. so $a_{2}>b$ for $k>1$. now we change $a$ and $b$ and get infinitely many pairs $(a,b)$. if we want to we could name them exactly... b) we want to show that all positive numbers greater than 1 are possible. so we set $a=k$ and $b=k^{2}-1$ where $k\in\mathbb{N}>1$: \[\frac{a^{2}+b^{2}-1}{ab}=\frac{k^{2}+k^{4}-2k^{2}+1-1}{k(k^{2}-1)}=\frac{k^{2}(k^{2}-1)}{k(k^{2}-1}=k\] $a$ and $b$ are greater than $1$ so $a^{2}+b^{2}-1>ab$ so we get $k>1$. Naphthalin

Actually your solution to b) implies a)... so why did you write a) too?^^

Naphthalin wrote: a) the equation is symmetric in $ a$ and $ b$. $ (2,3)$ is a solution with $ k = 2$. by vieta-jumping we get for the 2 zeroes $ a_{1}$ and $ a_{2}$ of $ a^{2} + b^{2} - 1 = kab$ the equations $ a_{1} + a_{2} = kb$ and $ a_{1}a_{2} = b^{2} - 1$. so $ a_{2} > b$ for $ k > 1$. now we change $ a$ and $ b$ and get infinitely many pairs $ (a,b)$. if we want to we could name them exactly... how does the above logic imply infinitely many solutions How exactly do we name them?? could anyone give a more analytical method. i.e without having to guess for the proper forms. something on the lines of solving the following equations \[ \frac{{\left( {a + b - 1} \right)\left( {a + b + 1} \right)}} {{ab}} = n + 2\]

Solution attached.

Attachments:
Brahmagupta.pdf (48kb)

(1) When $b=a+1$, $a^2 + b^2 - 1 = a^2 + (a+1)^2 - 1 = 2a(a+1)$, which is divided by $ab=a(a+1)$. FelixD wrote: Actually your solution to b) implies a)... so why did you write a) too?^^ (2) This problem can be solved by using Vieta-jumping method conversely. $\frac{a^2 + b^2 - 1}{ab} = m$ is equal to $a^2 - mab + b^2 - 1 = 0$. WLOG $a>b$, we can make another positive integer solution $(a', b)$ by setting $a'=mb-a$. By vieta theorem, $a \times a' = b^2 - 1 < a^2$; thus $a'<a$. This procedure lasts until $b=1$. Therefore the integer solution which gives the smallest sum is $(m,1)$. Then, the solution of the previous step would be $(a',b')$ which satisfies $m=mb'-a'$ and $1=b'$, or $a'=m^2 - 1$ and $b'=m$.

arkanm wrote: The first equation implies that $v_2$ is an integer, the second that $v_2>0$ since $A>B>1$. Thus, $(B,v_2)\in S$. Furthermore, $a+b$ is minimal at $(A,B)$... I assume that your definition of $S$ also requires $a,b > 1$ since you're using the fact that $A > B > 1$. In that case, this is wrong: you need $v_2 > 1$ to get the pair in $S$. In fact, $(a,b) = (8,3)$ yields $k=3$ if I'm not mistaken, so the claim you're trying to prove was wrong to begin with (and I just debugged it by plugging in $(a,b) = (8,3)$ and seeing what it Vieta jumped to). See the second post in the thread.

Naphthalin wrote: b) we want to show that all positive numbers greater than 1 are possible. so we set $a=k$ and $b=k^{2}-1$ where $k\in\mathbb{N}>1$: \[\frac{a^{2}+b^{2}-1}{ab}=\frac{k^{2}+k^{4}-2k^{2}+1-1}{k(k^{2}-1)}=\frac{k^{2}(k^{2}-1)}{k(k^{2}-1}=k\]$a$ and $b$ are greater than $1$ so $a^{2}+b^{2}-1>ab$ so we get $k>1$. Naphthalin What is the motivation for this construction?

mxgo wrote: Naphthalin wrote: b) we want to show that all positive numbers greater than 1 are possible. so we set $a=k$ and $b=k^{2}-1$ where $k\in\mathbb{N}>1$: \[\frac{a^{2}+b^{2}-1}{ab}=\frac{k^{2}+k^{4}-2k^{2}+1-1}{k(k^{2}-1)}=\frac{k^{2}(k^{2}-1)}{k(k^{2}-1}=k\]$a$ and $b$ are greater than $1$ so $a^{2}+b^{2}-1>ab$ so we get $k>1$. Naphthalin What is the motivation for this construction? The vieta jumping naphthalin did in part a) can be reversed until the equation produces the pair (a,1). This is clearly invalid, but then you note that roots $a_1, a_2$ satisfy $a_1 + a_2 = kb$ and $a_1 \cdot a_2 = b^2-1$. Letting $a_1$ = 1 implies that k = b. Moreover, we must be able to do this due to the "finite descent" the vieta jumping invoked. Since b can be varied over all integers, this implies that all k > 1 work. Tracing back our steps provides the solution a = k and b = $k^2-1$.

mxgo wrote: What is the motivation for this construction? I got the same construction, so here is my thought process: With no idea on how to start with (b), I started with (a). Naturally, I first set $a=2$ to get $3b|b^2-8$. It is clear that $b=3$ works. Here is when it struck me- does every solution of the form $(a,a+1)$ work? Substitute it, and we get $$E=\frac{a^2+b^2-1}{ab}=\frac{a^2+a^2+2a+(1-1)}{a(a+1)}=\frac{2a(a+1)}{a(a+1)}=2$$That's how some wishful thinking can come in handy! For the part (b), I guessed that every natural number is possible, for which all that was needed was to make a (clever) construction. Naturally, I tried to extend the above construction to $(a,b)=(a,a+k)$. I tried this, but it did not work. What is it that is not working here, but worked for $k=1$? After some thought, I realized it is that the $1$ gets canceled by the (annoying) $-1$! So I rather tried $(a,b)=(a,ka-1)$. We then get $$E=\frac{a^2+(ka-1)^2-1}{a(ka-1)}=\frac{(k^2+1)a-2k}{ka-1}=k+\frac{a-k}{ka-1}$$So we want $ka-1|a-k$. The simplest thing that we can do is to make $a-k=0$ by setting $a=k$! Boom, we are done! We have the construction $(a,b)=(k,k^2-1)$.