Integer Solutions: $(x^2-y^2)^2=1+16y$

[anemone]

Find all integer solutions of the equation $(x^2-y^2)^2=1+16y$

 

[mente oscura]

Find all integer solutions of the equation $(x^2-y^2)^2=1+16y$

Hello.

Missing me "thank you" button, anemone

Spoiler

1º) [tex]3|y[/tex]

Demonstration:

1.1) [tex]If \ 3|x,y \rightarrow{}3|(16y+1) \cancel{=}3k \ / \forall{} k \in{\mathbb{Z}}[/tex]. Absurdity.

1.2) [tex]If \ 3|x \ or \ 3|y \ or \ 3 \cancel{| } \ x,y \rightarrow{} (x^2-y^2)^2 \equiv{1 } mod(3)[/tex]

Therefore:

[tex][(x^2-y^2)^2-1] \equiv{0 } mod(3)[/tex]

2º) [tex]y^2<(16y+1)[/tex]

[tex]If \ y^2 \ge{} (16y+1) \rightarrow{}(x^2-y^2)^2=(x-y)^2(x+y)^2 > 16y+1[/tex]

[tex]y^2<16y+1 \rightarrow{} y^2-16y-1<0[/tex]

[tex]y< \dfrac{16 \pm \sqrt{256+4}}{2} < 17[/tex]

3º) The values of "and" generate "perfect" squares in [tex]16y+1[/tex] , they are the result of the following succession:

3.1) [tex]3(12n^2-17n+3) \rightarrow{}y=3,60,189,...[/tex]

3.2) [tex]3(12n^2-7n+3) \rightarrow{}y=18,105,264,...[/tex]

3.3) [tex]3(12n^2+10n-11) \rightarrow{}y=33,138,315,...[/tex]

3.4) [tex]3(12n^2+n) \rightarrow{}y=39,150,333,...[/tex]

Conclusion:

Only the result OK us: y=3

Let's see if it fulfils the conditions:

[tex](x^2-y^2)^2=16y+1[/tex]

[tex](x^2-9)^2=49[/tex]

[tex]x^2-9=7[/tex]

[tex]x^2=16 \rightarrow{}x=4 \ or \ x=-4[/tex]

(Whew)Regards.

 

[Opalg]

[sp]I found solutions $(x,y) = (\pm4,3),\ (\pm4,5)$. I will look at mente oscura's solution to see if it can be adapted to show that these are the only solutions.[/sp]
Further thoughts:
[sp]The left side of the equation $(x^2-y^2)^2 = 16y+1$ is positive, so $y$ must be positive. Let $x-y = k$. Then the equation becomes $k^2(2y-k)^2 = 16y+1.$

Case 1: $k>0$. If two positive integers have a given sum (which in this case will be $2y$) then their product is minimised by taking one of them equal to $1$ and the other to be $2y-1$. So the minimum value of $k(2y-k)$ is $2y-1$, and therefore $k^2(2y-k)^2 \geqslant (2y-1)^2$. Thus $ (2y-1)^2 \leqslant 16y+1$. That simplifies to $y^2 \leqslant 5y$, so that $y \leqslant 5$.

Case 2: $k<0$. In this case, let $r=-k$, so that the equation becomes $r^2(2y+r)^2 = 16y+1$. This time, $r$ is a positive integer, so that $r(2y+r) > 2y$ and hence $r^2(2y+r)^2 > 4y^2$. Thus $4y^2 < 16y+1$ from which it follows that $y\leqslant4.$

This shows that there are no solutions with $y>5$.

BUT ...

in a PM, anemone pointed out that I overlooked the solutions $(x,y) = (\pm1,0)$. So altogether there are six solutions $(\pm1,0), (\pm4,3), (\pm4,5)$.[/sp]

 

[mente oscura]

Hello.

Missing me "thank you" button, anemone

Spoiler

1º) [tex]3|y[/tex]

Demonstration:

1.1) [tex]If \ 3|x,y \rightarrow{}3|(16y+1) \cancel{=}3k \ / \forall{} k \in{\mathbb{Z}}[/tex]. Absurdity.

1.2) [tex]If \ 3|x \ or \ 3|y \ or \ [/tex][tex]3 \cancel{| } \ x,y[/tex][tex] \rightarrow{} (x^2-y^2)^2 \equiv{1 } mod(3)[/tex]

Hello.

Opalg, actually, I have an error.(Headbang)

Spoiler

[tex]3 \cancel{| } \ x,y[/tex] \rightarrow{} (x^2-y^2)^2 \equiv{0 } mod(3)[/tex]

It also fulfils the conditions:

[tex]x= \pm 4 \ and \ y=5[/tex]

Regards.

- - - Updated - - -

[sp]I found solutions $(x,y) = (\pm4,3),\ (\pm4,5)$. I will look at mente oscura's solution to see if it can be adapted to show that these are the only solutions.[/sp]
Further thoughts:
[sp]The left side of the equation $(x^2-y^2)^2 = 16y+1$ is positive, so $y$ must be positive. Let $x-y = k$. Then the equation becomes $k^2(2y-k)^2 = 16y+1.$

Case 1: $k>0$. If two positive integers have a given sum (which in this case will be $2y$) then their product is minimised by taking one of them equal to $1$ and the other to be $2y-1$. So the minimum value of $k(2y-k)$ is $2y-1$, and therefore $k^2(2y-k)^2 \geqslant (2y-1)^2$. Thus $ (2y-1)^2 \leqslant 16y+1$. That simplifies to $y^2 \leqslant 5y$, so that $y \leqslant 5$.

Case 2: $k<0$. In this case, let $r=-k$, so that the equation becomes $r^2(2y+r)^2 = 16y+1$. This time, $r$ is a positive integer, so that $r(2y+r) > 2y$ and hence $r^2(2y+r)^2 > 4y^2$. Thus $4y^2 < 16y+1$ from which it follows that $y\leqslant4.$

This shows that there are no solutions with $y>5$.

BUT ...

in a PM, anemone pointed out that I overlooked the solutions $(x,y) = (\pm1,0)$. So altogether there are six solutions $(\pm1,0), (\pm4,3), (\pm4,5)$.[/sp]

Correct.

 

[anemone]

[sp]I found solutions $(x,y) = (\pm4,3),\ (\pm4,5)$. I will look at mente oscura's solution to see if it can be adapted to show that these are the only solutions.[/sp]
Further thoughts:
[sp]The left side of the equation $(x^2-y^2)^2 = 16y+1$ is positive, so $y$ must be positive. Let $x-y = k$. Then the equation becomes $k^2(2y-k)^2 = 16y+1.$

Case 1: $k>0$. If two positive integers have a given sum (which in this case will be $2y$) then their product is minimised by taking one of them equal to $1$ and the other to be $2y-1$. So the minimum value of $k(2y-k)$ is $2y-1$, and therefore $k^2(2y-k)^2 \geqslant (2y-1)^2$. Thus $ (2y-1)^2 \leqslant 16y+1$. That simplifies to $y^2 \leqslant 5y$, so that $y \leqslant 5$.

Case 2: $k<0$. In this case, let $r=-k$, so that the equation becomes $r^2(2y+r)^2 = 16y+1$. This time, $r$ is a positive integer, so that $r(2y+r) > 2y$ and hence $r^2(2y+r)^2 > 4y^2$. Thus $4y^2 < 16y+1$ from which it follows that $y\leqslant4.$

This shows that there are no solutions with $y>5$.

BUT ...

in a PM, anemone pointed out that I overlooked the solutions $(x,y) = (\pm1,0)$. So altogether there are six solutions $(\pm1,0), (\pm4,3), (\pm4,5)$.[/sp]

Thank you Opalg for your elegant,neat and easy-to-follow solution! Well done!:)

Hello.

Opalg, actually, I have an error.(Headbang)

Spoiler

[tex]3 \cancel{| } \ x,y[/tex] \rightarrow{} (x^2-y^2)^2 \equiv{0 } mod(3)[/tex]

It also fulfils the conditions:

[tex]x= \pm 4 \ and \ y=5[/tex]

Regards.

- - - Updated - - -
Correct.

Don't worry about it, mente oscura! We live and learn!