This doesn't seem too likely

Lemma 1:

Suppose a certain pair of integers $(a,b)$ is a solution to the equation $n = \frac{a^2+b^2}{ab-1}$ . Then I say the pair $(b, \frac{b^2+n}{a})$ is also an integer solution.

Proof:

Consider the equation $x^2 - nbx + (b^2+n) = 0$ . By assumption, $a$ is one of the roots. Therefore, by Vieta's formulas, the other root is $\frac{b^2+n}{a}$ . By substituting the formula for $n$ , this equals $\frac{a^2+ab^3}{a(ab-1)}$ . We already know that the numerator is divisible by $ab-1$ (since $n$ is an integer), and it's clearly also divisible by $a$ . Since $gcd(a,ab-1) = 1$ , then the numerator is also divisible by $a(ab-1)$ . Hence the new root is an integer. Hence, $(b, \frac{b^2+n}{a})$ is also an integer solution. to the equation, $\boxed{ }$

Lemma 2:

If $a > b \ge 3$ , then $\frac{b^2+n}{a} < b$ .

Proof:

Since $b \ge 3$ , then $\frac{2b}{b^2-1} < 1$ . Therefore, $\frac{b^3+b}{b^2-1} = b + \frac{2b}{b^2-1} < b +1 \le a$ $b^3+b < ab^2-a$ $a+b^3 < ab^2-b$ $\frac{a+b^3}{ab-1} < b$ $\frac{b^2+n}{a} < b$ Hence $\frac{b^2+n}{a} < b$ . $\boxed{ }$

Now, let $(a,b)$ be the solution to $n = \frac{a^2+b^2}{ab-1}$ such that the value of $b$ is minimized (but nonnegative). Then, by Lemmas 1 and 2, we must have $0 \le b \le 2$ , otherwise it is possible to get a smaller value of $b$ .

If $b = 0$ , then $n = -a^2 \le 0$ .

If $b = 1$ , then $n = \frac{a^2+1}{a-1} = a+1 + \frac{2}{a-1}$ . Then $a = 0,2,3$ and therefore $n = -1, 5,5$ respectively.

If $b = 2$ , then we have $n = \frac{a^2 + 4}{2a-1}$ . Thus $4n = \frac{4a^2+16}{2a-1} = 2a+1+\frac{17}{2a-1}$ . Hence, $2a-1$ is a factor of 17. Therefore $a = 1, 9$ , and therefore $n = 5$ .

Therefore, the maximum possible value of $n$ in any of these cases is 5 (and we don't even need the restriction that $n \le 1000$ ). Hence the answer is $\boxed{5}$ .

Begin by moving everything on one side of the equation

$a^2-nab+n+b^2$

Use the quadratic equation to solve for a.

$a=\dfrac{nb\pm \sqrt{n^2b^2-4n-4b^2}}{2}$

Now we need to check integer solutions for the radical.

Note: if the first term of a square root of a second degree polynomial is a perfect square, then the sum of other terms must be a difference of squares.

Thus we can set our radical equal to $(nab+p)$ .

$(nab+p)^2=n^2a^2b^2-4n-4b^2$

$2nabp+p^2=-4n-4b$

$n=\dfrac{-(p^2+4b)}{2abp+4}$

We will use induction to find cases from here

for n to be an integer, p must be even.

since a,b are positive, to maximize n; p must be negative

to maximize (a,b), |p| must be minimized.

to maximize n; b must be maximized

to maximize b; a must be minimized

Thus plugging in p=-2, a=1 we get

$n=\dfrac{-4(b^2+1)}{-4(b-1)}$

$n=\dfrac{b^2+1}{b-1}$

This obviously doesn't have any solutions for b>3 since $b^2+1\equiv 2 \pmod{b-1}$

EDIT (Thanks to Sir Calvin):

The proof for this comes from the partial fraction decomposition of the expression above to $n=b+1+\dfrac{2}{b-1}$ , for which the last term is a non-integer for b>3.

Thus plugging in a and b into our original equation, $\boxed{n=5}$

[This is not a complete solution]

The case when $ab = 0$ is easily dealt with, and we have $n \leq 0$ .

If $a$ and $b$ are positive integers such that $\frac{ a^2 + b^2 } { ab-1 } = k$ is an integer, then $k = 5$ .

If you can prove this, I would appreciate if you can add your solution to the Vieta Root Jumping Wiki page.

Notice the symmetry: If $(a,\:b)$ is a solution to any given $n$ , so is $(b,\:a)$ . We only need to consider $0\leq a\leq b$

• Check some special cases with partial fraction decomposition (PFD). We must ignore $\red{a=b=1}$ , or we would divide by zero: $\begin{aligned} 0 &= a\leq b: &&& n &= -b^2\overset{!}{\in}\mathbb{Z} &&& \Rightarrow &&&& b &\geq 0, & n&\leq 0 \\[1em] 1 &= a\red{<} b: &&& n &= \frac{b^2+1}{b-1}=b+1+\frac{2}{b-1}\overset{!}{\in}\mathbb{Z} &&&\Rightarrow &&&& b&\in\{2;\:3\}, &n&=5\\[1em] 2 &\leq a=b: &&& n &= \frac{2a^2}{a^2-1}=2 + \frac{2}{a^2-1}\not\in\mathbb{Z}\\[1em] 2 &\leq a=b-1: &&& n &= \frac{a^2+(a+1)^2}{a(a+1)-1}=2 + \frac{3}{a(a+1)-1}\not\in\mathbb{Z}\\[1em] \end{aligned}$ In the last two cases, the denominator of the remainder is greater than the numerator and cannot generate an integer. The only remaining case we still need to check is $2\leq a\leq b-2$ $$

• Claim: Let $2\leq a\leq b-2$ . If $(a,\:b)$ is a solution to any given $n$ , so is $(a',\:b'):=(na-b,\:a)$ with $0<a'<a$ . $$ Proof: If $(a,\:b)$ is such a solution, then $n>0$ . We can write $b$ as a root of a polynomial $P(x)$ : $\begin{aligned} n&=\frac{a^2+b^2}{ab-1}\in\mathbb{Z} &&&\Rightarrow &&&& 0&=a^2+b^2-n(ab-1)=b^2-nab + a^2+n=:P(b) \end{aligned}$ With Vieta Root Jumping , we can find a second root $b^*$ of $P(x)$ : $\begin{aligned}b^*+b & = na &\Rightarrow &&a'&=b^*= na-b\in\mathbb{Z}, &&&&& a' &=b^*= \frac{a^2+n}{b}>0+0=0 \end{aligned}$ The other inequality $a'<a$ is a bit harder to prove. Insert the definition of $n$ into $a'$ and get $\begin{aligned} a'&=\frac{a^2+b^2}{ab-1}\cdot a-b=\frac{a^3+b}{ab-1}\underset{(*)}{\leq}\frac{a^3+(a+2)}{a(a+2)-1}\underset{\text{PFD}}{=}a-2+\frac{6a}{a(a+2)-1}\underset{(**)}{\leq} a-2+\frac{12}{7}<a \end{aligned}$ In $(*)$ , we use that the expression is decreasing in $b\geq a+2$ and $a\geq 2$ . In $(**)$ the remainder is decreasing in $a\geq 2\quad\blacksquare$ $$

• We still consider $2\leq a\leq b-2$ . For any solution $(a,\:b)$ to a given $n$ , we can repeat the last step to generate a sequence of solutions with decreasing $a'>0$ . After a finite number of steps, we have $a'=1$ , the second of our special cases! Every non-negative solution can be reduced to one of the special cases, so the maximum value is $n=\boxed{5}$

Rem.: We even showed that $n\in\{0;\:5\}$ are the only non-negative solutions to the problem! Both fundamental solutions with $a=1$ and $b\in\{2;\:3\}$ generate a countable sequence of solutions with $n=5$ . That was fun :)

Every number can be of the form 3k,3k+1,3k+2.

So using this we can find maximum value of n to be 5.

Sorry this is a wrong way to get answer.

The reason why I am not deleting this solution is there may be some one who thought the same like me so this may be helpful.😁

3 tries available despite incorrect answer. Searched up to 3 millions for n up to 1000 and 30 millions for n up to 100 via another modified parameters. Found that (n^2 - 4) b^2 and 4 n are to split after a formation for n = 5; such split makes possibility of formation of perfect square getting further; though, I consider my answer as a guess. There are many pairs of (a, b) along for n = 5 found by initial routine before the modification as follows. By rule of beauty, the most possible situation for search beyond 3 millions for n up to 1000 would not produce other wanted finding.

PROGRAM Largest_Integer;

USES CRT;

VAR

a, b, d, n, p: EXTENDED;

BEGIN

CLRSCR;

 n:=1000.0;

 REPEAT

     b:=2.0;

     REPEAT

           p:=b*n;

           d:=p*p-4.0*(b*b+n);

           IF d>=0 THEN

           BEGIN

                a:=0.5*(p+SQRT(d));

                IF (FRAC(a)=0.0) AND (n<>5.0) THEN

                BEGIN

                     WRITELN('Finished.');

                     WRITELN(n:1:0, '   ', a:1:6, '   ', b:1:0);

                     WRITE((a*a+b*b)/(a*b-1.0):1:18);

                     READLN;

                     EXIT

                END

           END;

           IF b=2999999.0 THEN

              WRITELN(n:1:0, '   ', a:1:6, '   ', b:1:0);

           b:=b+1.0

     UNTIL b=3000000.0;

     n:=n-1.0

 UNTIL n=2

END.

Consider the quadratic equation, x^2 - nbx + b^2 + n = 0. We know that one solution is a, and the other is c:= (b^2 + n)/a. From the equation on the problem, we let a = 1. So that n = (1+b^2)/(b-1) = (2+ b^2 - 1)/(b-1). Since b is a positive integer, then b-1 is also a positive integer. Since b^2 -1 is always divisible by b-1, and n is integer, then 2 must be divisible by b-1. From this we have b= 2 or b= 3. Since a=1 and b=2 then n=5. And since a=1 and b=3, we also have n=5. So the maximum of n is 5.

$We\quad may\quad use\quad induction\quad as\quad follws\\ 1)let\quad a=1\quad b=3\quad Therefore\quad n=5\\ 2)let\quad a=2\quad b=9\quad therefore\quad n=5\\ .\\ .\\ .\\ .\\ means\quad bw\quad every\quad 10\quad nos\\ like1-10,11-20.....991-1000\\ its\quad will\quad give\quad one\quad value\quad as\quad "'5""\\ \\ (I\quad know\quad its\quad just\quad the\quad triock\quad we\quad can\\ use\quad in\quad MCQs)\\ never\quad include\quad this\quad in\quad proofs$