Number Theory And Vieta Jumping

Question

$\textbf{Question:}$Find all positive integers $a, b$ such that the expression $$\frac{a^2+b^2+1}{ab-1}$$ is an integer.

$$$$As the expression is symmetric in $a, b$, so let $a \geq b$. It is easy to check the cases when $a=b+k$ where $0 \leq k \leq 2$, so let $a \geq b+3$. Suppose $a, b$ satisfies the given condition with $a+b$ minimized. Let us write $$\frac{a^2+b^2+1}{ab-1}=k$$ then our equation becomes $$a^2-kab+b^2+k+1=0$$. Now this equation has a root $a$ and from the Viet's formula we get the second root $$a'=kb-a$$ and $$a'=\frac{b^2+k+1}{a}$$. Now the first equation shows that $a'$ is an integer and the second equation shows that $a'$ is a positive integer. Now to show that $a' < a$ we have to prove that $$a'=\frac{b^2+k+1}{a} < a$$. Now consider $$a^2-(b^2+k+1)=a^2-b^2-1-k=a^2-b^2-1-(\frac{a^2+b^2+1}{ab-1})=\frac{a^3b-ab^3-ab-a^2+b^2+1-(a^2+b^2+1)}{ab-1}=\frac{a(a^2b-b^3-b-2a)}{ab-1}$$. Now as $a \geq b+3$ so we have $$a^2b-2a=a(ab-2) \geq a((b+3)(b)-2)=a(b^2+3b-2) \geq a(b^2+1) > b(b^2+1) > b^3+b$$ and hence we get $\frac{a(a^2b-b^3-b-2a)}{ab-1}>0$ and hence we have $a' < a$ and hence $a'+b < a+b$ contradicting the minimality of $a+b$. So there are no solutions. $$$$Is My Proof Correct??

Answer 1

Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica code:

In[1]:=ParallelTable[
  If[IntegerQ[(a^2 + b^2 + 1)/(a*b - 1)], {a, 
    b, (a^2 + b^2 + 1)/(a*b - 1)}, Nothing], {a, 2, 10000}, {b, 2, 
   10000}] //. {} -> Nothing

Running the code gives:

Out[1]={{{2, 2, 3}, {2, 4, 3}, {2, 11, 6}}, {{4, 2, 3}, {4, 10, 3}, {4, 23, 
   6}}, {{10, 4, 3}, {10, 26, 3}}, {{11, 2, 6}, {11, 64, 6}}, {{23, 4,
    6}, {23, 134, 6}}, {{26, 10, 3}, {26, 68, 3}}, {{64, 11, 6}, {64, 
   373, 6}}, {{68, 26, 3}, {68, 178, 3}}, {{134, 23, 6}, {134, 781, 
   6}}, {{178, 68, 3}, {178, 466, 3}}, {{373, 64, 6}, {373, 2174, 
   6}}, {{466, 178, 3}, {466, 1220, 3}}, {{781, 134, 6}, {781, 4552, 
   6}}, {{1220, 466, 3}, {1220, 3194, 3}}, {{2174, 373, 6}}, {{3194, 
   1220, 3}, {3194, 8362, 3}}, {{4552, 781, 6}}, {{8362, 3194, 3}}}

Where the code looks for integer solutions to your problem given as $\left(\text{a},\text{b},\frac{\text{a}^2+\text{b}^2+1}{\text{a}\text{b}-1}\right)$.

So, we can see that your claim that there are no solutions is certainly not true for $2\le\text{a}\le10^4$ and $2\le\text{b}\le10^4$.

Answer 2

There are infinitely many solutions with $k=3$ and $k=6.$ For 3 the generating point is $(2,2)$ For $k=6$ the points are $(2,1)$ and $(1,2)$ More solutions can be found in either case by Vieta jumping.

Ruling out larger $k$ is by inequalities.

See how, as soon as $k \geq 7,$ the arc of the hyperbola passes through the interior of the square $1 < x < 2, $ $ 1 < y < 2.$ The intersections with the bounding lines $y = \frac{k}{2} x$ and $x = \frac{k}{2} y$ have one coordinate strictly below $1.$ As a result, there are no integer points on the arc between the bounding lines, therefore no integer points in the first quadrant at all. The method is described in HURWITZ 1907