# Solutions to the diophantine equation $6x^2 - 6x - y^2 + y=0$?

Are there any positive integer solutions to the diophantine equation in the title other than $(1,1)$?

This equation looks easy enough so it could be that there is some simple argument that shows that there are or that there are not other solutions but my brain does not work good enough right now. Thank you for the help.

Bu completing the square in both $x$'and $y$ we get:

$(2y-1)^2-6(2x-1)^2=-5$.

This is a Pell-type equation and given one solution, it will have infinity many. From the solution (1,1) we get a family:

$(x,y)=(1,1),(4,9),(35,85),etc.$

There is a second "seed" at (2,4) (seen in a different answer to this question), thus:

$(x,y)=(2,4),(15,36),(144,352),etc.$

• for each of these, $x_{n+2} = 10 x_{n+1} - x_n -4$ and $y_{n+2} = 10 y_{n+1} - y_n -4$ Apr 12 '16 at 2:57

I'm going to switch the letters, $$x^2 - x = 6 y^2 - 6 y.$$ $$4x^2 - 4x = 24 y^2 - 24 y$$ $$4x^2 - 4x+1 = 24 y^2 - 24 y + 6 - 5$$ $$(2x-1)^2 = 6 (2y-1)^2 - 5$$ $$(2x-1)^2 - 6 (2y-1)^2 = - 5$$ This is a Pell type equation, $u^2 - 6 v^2 = -5.$

  5^2 - 6 2^2 = 1

u^2 - 6 v^2 = -5

Mon Apr 11 19:07:34 PDT 2016

u:  1  v:  1 ratio: 1  SEED
u:  7  v:  3 ratio: 0.4285714285714285  SEED
u:  17  v:  7 ratio: 0.4117647058823529
u:  71  v:  29 ratio: 0.4084507042253521
u:  169  v:  69 ratio: 0.408284023668639
u:  703  v:  287 ratio: 0.4082503556187767
u:  1673  v:  683 ratio: 0.4082486551105798
u:  6959  v:  2841 ratio: 0.4082483115390142
u:  16561  v:  6761 ratio: 0.4082482941851337
u:  68887  v:  28123 ratio: 0.4082482906789379
u:  163937  v:  66927 ratio: 0.4082482905018391
u:  681911  v:  278389 ratio: 0.4082482904660579
u:  1622809  v:  662509 ratio: 0.4082482904642505
u:  6750223  v:  2755767 ratio: 0.4082482904638854
u:  16064153  v:  6558163 ratio: 0.408248290463867

Mon Apr 11 19:07:54 PDT 2016

u^2 - 6 v^2 = -5


For each $(u,v)$ pair, you get another from $$(u,v) \mapsto (5u+12v, 2u+5v)$$

Both are always odd, take $$x = \frac{u+1}{2}, \; \; y = \frac{v+1}{2}$$

I got one pair $(x,y)$ as $(2,4)$,