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- #1

#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$2a^2+a+1=y^2$

Is it possible to solve this equation systematically, or will it involve guesswork?

it seemed easy ,but....

you may check it using program (I have done this )

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

$2a^2+a+1=y^2$

Is it possible to solve this equation systematically, or will it involve guesswork?

it seemed easy ,but....

you may check it using program (I have done this )

- Thread starter
- #2

- Jan 25, 2013

- 1,225

in this case the numbers of solutions will be infinite or finite ?

can we prove it ?

up to now we know :

if y odd ,a must be even

if y even ,a must be odd

can we prove it ?

up to now we know :

if y odd ,a must be even

if y even ,a must be odd

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- #3

- Feb 7, 2012

- 2,792

If $2a^2 + a + 1 = y^2$ then (completing the square) $2\bigl(a+\frac14\bigr)^2 + \frac78 = y^2$. Multiply both sides by $8$ and rearrange, to get $(4a+1)^2 - 2(2y)^2 = 7.$ This is a Diophantine equation of the form $X^2 - 2Y^2 = $ const. It has infinitely many solutions, starting with the obvious ones $(a,y) = (0,\pm1)$ and $(a,y) = (1,\pm2)$. It can be solved by using the techniques for Pell's equation (we may have to call in our resident expert MarkFL at this point). You can find, for example, that if $(a,y)$ is a solution then so are $(17a+4\pm 12y,\,24a+6\pm 17y)$. That gives solutions such as $(-3,4)$, $(-8,11)$, $(16,23)$, $(45,64)$, $(552,781)$, $\ldots$.

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- #5

- Feb 7, 2012

- 2,792

Stop making such nice remarks about Mark:(we may have to call in our resident expert MarkFL at this point)....

it'll go to his head, and he'll stop buying the rounds