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
Solution 1:
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.
Solution 2:
I like it.
1225 m-n 6 105 99 1225 = 5^2 * 7^2
9801 m-n 10 495 485 9801 = 3^4 * 11^2
38025 m-n 14 1365 1351 38025 = 3^2 * 5^2 * 13^2
104329 m-n 18 2907 2889 104329 = 17^2 * 19^2
1413721 m-n 204 3567 3363 1413721 = 29^2 * 41^2
233289 m-n 22 5313 5291 233289 = 3^2 * 7^2 * 23^2
455625 m-n 26 8775 8749 455625 = 3^6 * 5^4
808201 m-n 30 13485 13455 808201 = 29^2 * 31^2
1334025 m-n 34 19635 19601 1334025 = 3^2 * 5^2 * 7^2 * 11^2
2082249 m-n 38 27417 27379 2082249 = 3^2 * 13^2 * 37^2
3108169 m-n 42 37023 36981 3108169 = 41^2 * 43^2
The main sequence has $$ m = 32 w^3 + 48 w^2 + 22 w + 3, $$ $$ n = 32 w^3 + 48 w^2 + 18 w + 1, $$ $$ m^2 - n^2 + 1 = \left( (4w+1)(4w+3) \right)^2 $$ $$ n^2 - 1 = \; 4 \, w \; (w+1) \; \left( (4w+1)(4w+3) \right)^2 $$ This does NOT include
1413721 m-n 204 3567 3363 1413721 = 29^2 * 41^2