Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$

Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$. What are the possible values of $$\frac{x^2+y^2-1}{xy}$$?

I have discovered infinitely many pairs. They are $(n, n+1)$. This solution works for all $n$. For all of them the value of $$\frac{x^2+y^2-1}{xy}=2$$

But I want to know if there are any other such solutions. Also what are all the possible values. I have discovered 2. There may be many such values. Are they finite? Can we prove their finite or infinite existence?

Any help would be appreciated.

• There is an article in the Excalibur Journal in Hong Kong. Check it out back issue that wrote about this. – DeepSea Mar 20 '16 at 10:07
• @Kf-Sansoo I googled it and can't seem to find it. – TheRandomGuy Mar 20 '16 at 10:10
• Its called Math Excalibur free download. I visited this site reading it dozen times in past. – DeepSea Mar 20 '16 at 10:11
• You cannot have spent much time looking! There are oodles of them. Eg: $8^2+3^2-1=3 \times8 \times 3,15^2+4^2-1=4 \times15\times4$ – almagest Mar 20 '16 at 10:59
• Oh, maybe you meant you had discovered two others. Well here is a fourth $(24^2+5^2-1)/(24\times5)=5$, and $(35^2+6^2-1)(35\times6)=6$. – almagest Mar 20 '16 at 11:03

Note that the usual problem, and quite difficult if you do not know what to expect, is slightly different, https://en.wikipedia.org/wiki/Vieta_jumping#Example_2

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Define integer $k \geq 3$ as $$k = \frac{x^2+y^2-1}{xy}$$ We are solving $$x^2 - kxy + y^2 = 1.$$ The discriminant of the quadratic form is $\Delta = k^2 - 4$ which is positive but not a square. As a result, if we can solve $\tau^2 - \Delta \sigma^2$ we can construct the generator of the (oriented) automorphism group of the form. Now, $k^2 - \Delta = 4,$ so $\tau = \pm k, \sigma = \pm 1.$

Given a form (positive nonsquare discriminant) of coefficients $\langle A,B,C\rangle$ we can take the generator (in $SL_2 \mathbb Z$) as $$\left( \begin{array}{rr} \frac{\tau - B \sigma}{2} && -C \sigma \\ A \sigma && \frac{\tau + B \sigma}{2} \end{array} \right)$$ With $\langle A,B,C\rangle = \langle 1,-k,1\rangle$ we get $$U = \left( \begin{array}{rr} k && -1 \\ 1 && 0 \end{array} \right)$$ and will stick with this one, as convenient for keeping $x \geq y \geq 0.$ The transformation from one solution $(x,y)$ to the next is matrix multiplication with the column vector $(x,y)^T$ on the right. That is, $$\color{blue}{ (x,y) \mapsto (kx-y, x)}.$$ The first few solutions with $k \geq 3$ are $$(1,0),$$ $$(k,1),$$ $$(k^2 - 1,k),$$ $$(k^3 - 2k,k^2 - 1),$$ $$(k^4 - 3 k^2 + 1,k^3 - 2k).$$ Oh, as far as actual division, $(1,0)$ is not legal in the original fraction. Life is like that.

Let's try $k=3$ in the fraction: $(3,1)$ gives $9/3 = 3.$ $(8,3)$ gives $72/24 = 3.$

The final thing is Cayley-Hamilton. The matrix I named $U$ above satisfies $U^2 - k U + I = 0,$ or $U^2 = kU - I.$ As a result, both $x$ and $y$ satisfy linear recurrences, $$x_{j+2} = k x_{j+1} - x_j,$$ $$y_{j+2} = k y_{j+1} - y_j.$$ Again with $k=3,$we get $x$ in $$1, 3, 8, 21, 55, 144,$$ Note that these are every second Fibonacci number.

With $k=4,$we get $x$ in $$1, 4, 15, 56, 209, 780,$$

With $k=5,$we get $x$ in $$1, 5, 24, 115, 551, 2640,$$

With your $k=2$ you have $x^2 - 2xy + y^2 = 1,$ or $(x-y)^2 = 1,$ or $x-y = \pm 1.$

Every $n$ works.

$$\forall n\in\mathbb{Z}^+. \,\,\,\,n\cdot n \cdot (n^2-1) = n^2+ (n^2-1)^2 -1$$

So,

$$\forall n\in\mathbb{Z}^+.\,\,\,\,\ \frac{n^2 + (n^2-1)^2 -1}{n \cdot (n^2 -1)} = n$$