Showing that $m^2-n^2+1$ is a square

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

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.