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I am trying to find all integer solutions to the following equation: $$32x^2 - y^2 = 448$$

This is what I have tried so far:

The equation describes a hyperbola, and so I try the usual trick of intersecting the curve with a line of rational slope to find rational solutions first.

Knowing the point (4,8) satisfies the equation, I solve the following system: $$\left\{ \begin{array}{c} 32x^2 - y^2 = 448 \\ y = m(x - 4) + 8 \\ \end{array} \right.$$

After a bunch of algebra, I get: $$x = \frac{-4m^2+16m-128}{32-m^2}$$ $$y = \frac{8m^2-256m+256}{32-m^2}$$

Finally, substituting $m = \frac{u}{v}$, I get: $$x = \frac{-4u^2+16uv-128v^2}{32v^2-u^2}$$ $$y = \frac{8u^2-256uv+256v^2}{32v^2-u^2}$$

Cool, with any choice of $u$ and $v$, I get a rational solution.

But since cancelling the denominators does not work, I do not know how to continue to get integer solutions only.

Is this perhaps not the right way to go? Any help would be much appreciated.

Michael Rozenberg
  • 194,933
BizarreCake
  • 151

3 Answers3

3

$y^2$ divisible by $64$.

Let $y=8y_1$.

Thus, we have $$x^2-2y_1^2=14,$$ which says that $x$ divisible by $2$.

Let $x=2x_1$.

Thus, we need to solve $$2x_1^2-y_1^2=7,$$ which reduce to Pell.

https://en.wikipedia.org/wiki/Pell%27s_equation

Michael Rozenberg
  • 194,933
2

Four orbits under $$ x_{n+2} = 34 x_{n+1} - x_n, $$ $$ y_{n+2} = 34 y_{n+1} - y_n. $$

$$ (4,8); \; \; (92,520); \; \; (3124,17672); \; \; (106124,600328); $$ $$ (8,40); \; \; (256,1448); \; \; (8696,49192); \; \; (295408,1671080); $$ $$ (16,88); \; \; (536,3032); \; \; (18208,103000); \; \; (618536,3498968); $$ $$ (44,248); \; \; (1492,8440); \; \; (50684,286712); \; \; (1721764,9739768); $$

As sometimes happens, these can be combined into two orbits under $$ x_{n+2} = 6 x_{n+1} - x_n, $$ $$ y_{n+2} = 6 y_{n+1} - y_n. $$

$$ (4,8); \; \; (16,88); \; \; (92,520); \; \; (536,3032); \; \; (3124,17672); \; \;(18208,103000); \; \; (106124,600328); \; \; (618536,3498968);$$ $$ (8,40); \; \; (44,248); \; \;(256,1448); \; \;(1492,8440); \; \; (8696,49192); \; \; (50684,286712); \; \; (295408,1671080); \; \;(1721764,9739768); $$

My program calls them $w,v.$

==================================================

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:  
    17   96
    3   17
  Automorphism backwards:  
    17   -96
    -3   17

  17^2 - 32 3^2 = 1

 w^2 - 32 v^2 = -448

Wed Oct  4 07:13:21 PDT 2017

w:  8  v:  4 ratio: 2  SEED   KEEP +- 
w:  40  v:  8 ratio: 5  SEED   KEEP +- 
w:  88  v:  16 ratio: 5.5  SEED   BACK ONE STEP  -40 ,  8
w:  248  v:  44 ratio: 5.63636  SEED   BACK ONE STEP  -8 ,  4
w:  520  v:  92 ratio: 5.65217
w:  1448  v:  256 ratio: 5.65625
w:  3032  v:  536 ratio: 5.65672
w:  8440  v:  1492 ratio: 5.65684
w:  17672  v:  3124 ratio: 5.65685
w:  49192  v:  8696 ratio: 5.65685
w:  103000  v:  18208 ratio: 5.65685
w:  286712  v:  50684 ratio: 5.65685
w:  600328  v:  106124 ratio: 5.65685
w:  1671080  v:  295408 ratio: 5.65685
w:  3498968  v:  618536 ratio: 5.65685
w:  9739768  v:  1721764 ratio: 5.65685
w:  20393480  v:  3605092 ratio: 5.65685
w:  56767528  v:  10035176 ratio: 5.65685
w:  118861912  v:  21012016 ratio: 5.65685
w:  330865400  v:  58489292 ratio: 5.65685

Wed Oct  4 07:15:22 PDT 2017

 w^2 - 32 v^2 = -448

jagy@phobeusjunior:~$

============================

Will Jagy
  • 139,541
1

Note that you can simplify and substitute in the following way: $$\begin{array}{lll} y^2 = 2^5(x^2-14)&&\text{substitute } y = 2^2z \\ z^2 = 2(x^2-14)&&\text{substitute } z = 2a\\ 2a^2 = x^2-14&& \\ 2(a^2+7) = x^2&&\text{substitute } x = 2b\\ a^2+7 = 2b^2&&\text{substitute } a = 2c+1\\ 2(c^2+c + 2) = b^2&&\text{substitute } b = 2d\\ 2d^2 = c^2 + c + 2&&\\ 2(d-1)(d+1) = c(c+1)&&\\ \end{array}$$ so that $y = 16c+8$ and $x=4d$. This might work better for your approach.

Yannik
  • 628