Showing that $m^2-n^2+1$ is a square Prove that if $m,n$ are odd integers such that $m^2-n^2+1$ divides $n^2-1$ then $m^2-n^2+1$ is a square number. 
I know that a solution can be obtained from Vieta jumping, but it seems very different to any Vieta jumping problem I've seen.
To start, I chose $m=2a+1$ and and $n=2b+1$ which yields: $$ 4ka^2+4ka-4kb^2-4kb+k = b^2+b$$
Then suppose that $B$ is a solution, and $B_0$ is another solution. Then using Vieta jumping we get (with a bit of algebra) that $B+B_0 = -1$ and $B_0 = \frac {-k(2a+1)^2}{B(4k+1)}$.
But I'm not sure these final equalities are particularly helpful; I can't find any way to yield more solutions from them.  How can I solve the problem? A solution without Vieta jumping is probably also possible
 A: Under the assumption that the integer ratio is positive:
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LEMMA
Given integers $$  M \geq m > 0, $$ along with positive integers $x,y$ with
$$ x^2 - Mxy + y^2 = m.  $$
Then $m$ is a square.
PROOF.
First note that we cannot have integers $xy < 0$ with $ x^2 - Mxy + y^2 = m,  $ since then
 $ x^2 - Mxy + y^2 \geq 1 + M + 1 = M + 2 > m.$ If we have a solution with $x > 0$ and $xy \leq 0,$ it follows that $y=0.$ 
This is the Vieta jumping part, with some extra care about inequalities.
Case I: We begin with integers $$  y > x > 0 $$ and the stronger $$y > Mx.$$
Then we get a new solution by jumping
$$ (x,y) \mapsto (Mx - y,x). $$
However, the assumption $y > Mx$ means $Mx-y < 0,$ we cannot have a solution with one variable positive and the other negative. This case cannot occur.
Case II.  $y > x > 0$ and $y = Mx.$ But then $x^2 - Mxy + y^2 = x^2 - M^2 x^2 + M^2 x^2 = x^2.$ Therefore $x^2 = m$ which is a square.
Case III.
$$ y > x  $$ and
$$ y < Mx.  $$ We have
$$ x^2 - Mxy + y^2 > 0,  $$
$$ x^2 > Mxy - y^2 = y(Mx - y) > x(Mx-y), $$
$$ x > Mx - y > 0. $$
That is, the jump
$$ (x,y) \mapsto (Mx - y,x) $$
takes us from one ordered solution to another ordered solution while strictly decreasing $x+y.$
Within a finite number of such jumps we violate the conditions we were preserving; we reach a solution $(x,y)$ with $y \geq Mx,$ that is
$x > 0$ but $Mx-y \leq 0.$ Since $(Mx - y,x) $ is another solution we know that $Mx-y = 0.$ Therefore $x^2 = m$ and $m$ is a square.
Graph for $x^2 - 5xy + y^2 = 3$

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Suppose we have odd integers $m,n > 0$ such that
$$  \frac{n^2 - 1}{m^2 - n^2 + 1} = k > 0 $$
is an integer. Then 
$$ k+1 = \frac{m^2 }{m^2 - n^2 + 1}.  $$
Name $w = 1 + k,$ so
$$ w = \frac{m^2 }{m^2 - n^2 + 1}.  $$
We are sticking with positive $w$ so we may take $m \geq n >0.$ When we write
$$ m-n = 2x,  $$
$$ m+n = 2y,  $$
we are introducing positive variables. Then $m=x+y,$ $n = y - x,$ and
$$ w = \frac{x^2 + 2xy + y^2}{4xy+1}, $$
$$ x^2 + 2xy+ y^2 = 4wxy + w,  $$
$$ x^2 - (4w-2)xy + y^2 = w. $$
From the LEMMA, we find that $w$ is a square. From
$$ w = \frac{m^2 }{m^2 - n^2 + 1}  $$
we see that 
$$ m^2 - n^2 + 1  $$ is also a square.
A: Proof: First, consider $n^2 - 1 = (d^2 - 1)(m^2 - n^2 + 1)$, where $d \mid m $, $d$ is a positive divisor of $m$.
            It is easy to see that in this case $m^2 - n^2 + 1 = \left( \frac{m}d \right)^2 $. Then it suffices to show
            that these are the only possibilities of writing
            $$ n^2 - 1 = k(m^2 - n^2 + 1), \hspace{3em} \text{ (*)}$$
            where $ k $ must be of the form $ d^2 - 1$. 
Note by (*) we have that $ k $ is bounded by $m^2 - 1$.
            (Since $ k \leqslant n^2 -1 \leqslant m^2 - 1 $.) Also by (*) we notice that $k m^2 = (k+1)(n^2 - 1)$.
            $k$ must be even otherwise $ m $ cannot be odd.
            But $ k+1 \nmid k $, it implies we must have $ k + 1 \mid m^2 $, i.e.,  $ k $ must be of the form $ d^2 - 1 $. Q.E.D.

Edit:
Needs to fix the gap for the struck through implication.
A: I have produced the lemma that rules out negative ratios. It is to be applied after the business of wrting the sum and difference of the pair of odd integers as double new variables.  
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LEMMA
Given integers $$   m > 0, \; \; M > m+2, $$ there are no integers $x,y$ with
$$ x^2 - Mxy + y^2 = -m.  $$
PROOF
Calculus: $m+2 >  \sqrt{4m+4},$ since $(m+2)^2 = m^2 + 4m + 4,$ while $\left(  \sqrt{4m+4}  \right)^2 = 4m + 4.$ Therefore also
$$ M > \sqrt{4m+4} $$
We cannot have $xy < 0,$ as then $x^2 - M xy + y^2 \geq 2 + M > 0. $ It is also impossible to have $x=0$ or $y=0.$ From now on we take integers $x,y > 0.$
With $x^2 - Mxy + y^2 < 0,$ we get $0 < x^2 < Mxy - y^2 = y(Mx - y),$ so that $Mx - y > 0$ and $y < Mx.$ We also get
$x < My.$
The point on the hyperbola $ x^2 - Mxy + y^2 = -m  $ has both coordinates $x=y=t$ with $(2-M) t^2 = -m,$ $(M-2)t^2 = m,$ and
$$ t^2 = \frac{m}{M-2}. $$
We demanded $M > m+2$ so $M-2 > m,$ therefore $t < 1.$
 More important than first appears, that this point is inside the unit square. 
We now begin to use the viewpoint of Hurwitz (1907). All elementary, but probably not familiar. We are going to find integer solutions that minimize $x+y.$ If $2 y > M x,$ then $y > Mx-y.$ Therefore, when Vieta jumping, the new solution given by
$$ (x,y) \mapsto (Mx - y, x)  $$
gives a smaller $x+y$ value. Or, if $2x > My,$
$$ (x,y) \mapsto (y, My - x)  $$
gives a smaller $x+y$ value. We already established that we are guaranteed $My-x, Mx-y > 0.$
Therefore, if there are any integer solutions, the minimum of $x+y$ occurs under the Hurwitz conditions for a fundamental solution (Grundlösung), namely
$$ 2y \leq Mx  \; \; \; \; \mbox{AND} \; \; \; \;  2 x \leq My.  $$
We now just fiddle with calculus type stuff, that along the hyperbola arc bounded by the Hurwitz inequalities, either $x < 1$ or $y < 1,$ so that there cannot be any integer lattice points along the arc. We have already shown that the middle point of the arc lies at $(t,t)$ with $t < 1.$  We just need to confirm that the boundary points also have either small $x$ or small $y.$ Given $y = Mx/2,$ with
$$ x^2 - Mxy + y^2 = -m $$ becomes
$$ x^2 - \frac{M^2}{2} x^2 + \frac{M^2}{4} x^2 = -m,  $$
$$ x^2 \left( 1 - \frac{M^2}{4}  \right) = -m  $$
$$ x^2 = \frac{-m}{1 - \frac{M^2}{4}} =  \frac{m}{ \frac{M^2}{4} - 1} = \frac{4m}{M^2 - 4}.  $$
We already confirmed that $ M > \sqrt{4m+4}, $ so $M^2 > 4m+4$ and $M^2 - 4 > 4m.$ As a result, $ \frac{4m}{M^2 - 4} < 1.$ The intersection of the hyperbola with the Hurwitz boundary line $2y = Mx$ gives a point with $x < 1.$ Between this and the arc middle point, we always have $x < 1,$ so no integer points. Between the arc middle point and the other boundary point, we always have $y < 1.$ All together, there are no integer points in the bounded arc. There are no Hurwitz fundamental solutions. Therefore, there are no integer solutions at all.
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