Solve $(a^2-b^2)^2=1+16a$ for Integers

[anemone]

Solve the equation in the set of integers:

$(a^2-b^2)^2=1+16a$

 

[Dethrone]

Spoiler

$$(a^2-b^2)^2=1+16a$$
$$(a^2-b^2)^2-16a=1$$
$$(a^2-b^2-4\sqrt{a})(a^2-b^2+4\sqrt{a})=1$$

This implies that either both terms are 1, or both terms are negative 1. Trying for -1:
$$a^2-b^2-4\sqrt{a}=-1$$
$$a^2-b^2+4\sqrt{a}=-1$$

I'm not aware of any faster way of solving this, but let me know if there is :D
Add both equations:
$$2(a^2-b^2)=-2$$
$$(a+b)(a-b)=-1$$

This further implies that either both terms are negative, or both are positive. Solving both possible cases:
$a+b=1$
$a-b=-1$

and

$a+b=-1$
$a-b=1$

Gives us the integer solutions $(0, 1)$ and $(0, -1)$, respectively.

 

[Nono713]

Spoiler

$$(a^2-b^2)^2=1+16a$$
$$(a^2-b^2)^2-16a=1$$
$$(a^2-b^2-4\sqrt{a})(a^2-b^2+4\sqrt{a})=1$$

This implies that either both terms are 1, or both terms are negative 1. Trying for -1:
$$a^2-b^2-4\sqrt{a}=-1$$
$$a^2-b^2+4\sqrt{a}=-1$$

I'm not aware of any faster way of solving this, but let me know if there is :D
Add both equations:
$$2(a^2-b^2)=-2$$
$$(a+b)(a-b)=-1$$

This further implies that either both terms are negative, or both are positive. Solving both possible cases:
$a+b=1$
$a-b=-1$

and

$a+b=-1$
$a-b=1$

Gives us the integer solutions $(0, 1)$ and $(0, -1)$, respectively.

Spoiler

Your proof does not follow because $a^2 - b^2 \pm 4 \sqrt{a}$ are not necessarily integers.

 

[anemone]

Spoiler

$$(a^2-b^2)^2=1+16a$$
$$(a^2-b^2)^2-16a=1$$
$$(a^2-b^2-4\sqrt{a})(a^2-b^2+4\sqrt{a})=1$$

This implies that either both terms are 1, or both terms are negative 1. Trying for -1:
$$a^2-b^2-4\sqrt{a}=-1$$
$$a^2-b^2+4\sqrt{a}=-1$$

I'm not aware of any faster way of solving this, but let me know if there is :D
Add both equations:
$$2(a^2-b^2)=-2$$
$$(a+b)(a-b)=-1$$

This further implies that either both terms are negative, or both are positive. Solving both possible cases:
$a+b=1$
$a-b=-1$

and

$a+b=-1$
$a-b=1$

Gives us the integer solutions $(0, 1)$ and $(0, -1)$, respectively.

Hi Rido12, (Wave)

Thanks for participating and Bacterius is right, I'm sorry that your solution doesn't work out to a full solution :( but I applaud you for attempting it. (Yes)

 

[topsquark]

Solve the equation in the set of integers:

$(a^2-b^2)^2=1+16a$

[sp]I'm not going to write it out because it's all brute force and checking a lot of cases, but it can be done a modulo 4. If that's the only way I'll post, but I'd highly prefer a more elegant solution.)[/sp]

-Dan

 

[anemone]

[sp]I'm not going to write it out because it's all brute force and checking a lot of cases, but it can be done a modulo 4. If that's the only way I'll post, but I'd highly prefer a more elegant solution.)[/sp]

-Dan

Thanks, topsquark for your reply. As far as I can tell, the modular arithmetic method isn't the only way to solve for this problem but I am willing to wait for a day or two before I posting the solution I've at hand, just in case there are other members who have still not thrown in the towel yet and determined to solve it using another approach. Okay

 

[topsquark]

(sighs) I ran into a hiccup.

I just went through the whole thing again to try and streamline it and I ran into a problem in my original work. I can whittle it down to two possibilities, [math]a^2 \equiv 1,~b^2 \equiv 0~\text{mod(4)}[/math], which should clearly be not true, but proving it is nasty and frankly I don't know how to do it, or [math]a^2 \equiv 0,~b^2 \equiv 1~\text{mod(4)}[/math], which clearly generates the two solutions, but proving that these are the only ones is nasty and here again I don't know how to do it. Without my mistake it was reasonably simple, if long. Ah well.

Back to square one I'm afraid.

-Dan

Addendum: Just thought I'd mention, my mistake was an easy one to make. Taking [math]a^2 \equiv 1,~b^2 \equiv 0~\text{mod(4)}[/math] I originally took a = 2n + 1 and b = 2n. Of course, this isn't general. It should be a = 2n + 1 and b = 2m. When I fixed this and plugged the a = 2n + 1 into the original equation to find b(n) I got a horrible result for b which I don't know how to analyze in terms of integer solutions.

 

[Albert1]

Solve the equation in the set of integers:

Spoiler

$(a^2-b^2)^2=1+16a--(1)$

Spoiler

from (1) we get :
$a\geq 0$
and $1+16a$ is a perfect square
if $a=0,\,\,b=\pm 1$
if $a>0:$
$a^2-b^2=c^2=\sqrt {16a+1}---(2)$
$a,b,c \in N$
now consder $a,b,c$ as 3 side length of a
right angled tiangle ABC,then
$a,c $ must be odd ,and $b$ even
let :$a=m^2+n^2,b=2mn, c=m^2-n^2--(3)$
from (2)(3)$m=\pm 2,\,\ n=\pm 1$
and we may conclude the corresponding solutions:
$(a,b)=(0,\pm 1),(5,\pm 4)$

 

[kaliprasad]

Spoiler

from (1) we get :
$a\geq 0$
and $1+16a$ is a perfect square
if $a=0,\,\,b=\pm 1$
if $a>0:$
$a^2-b^2=c^2=\sqrt {16a+1}---(2)$
$a,b,c \in N$
now consder $a,b,c$ as 3 side length of a
right angled tiangle ABC,then
$a,c $ must be odd ,and $b$ even
let :$a=m^2+n^2,b=2mn, c=m^2-n^2--(3)$
from (2)(3)$m=\pm 2,\,\ n=\pm 1$
and we may conclude the corresponding solutions:
$(a,b)=(0,\pm 1),(5,\pm 4)$

this leaves out the possibility that

Spoiler

$a^2-b^2 = - c^2$

 

[anemone]

Solution of other:

Spoiler

It is evident that if $(a,\,b)$ is a solution of the equation, then $(a,\,-b)$ is also its solution. Hence, it's sufficient to consider $b \ge 0$.

Since the RHS is a non-negative, we could deduce that $1+16a\ge 0$. That is, $a$ takes positive integers value only. Therefore $(a^2-b^2)^2\ge 1$. This also implies $(a-b)^2 \ge 1$.

Hence, $(a^2-b^2)^2=(a+b)^2(a-b)^2\ge a^2 $.

From this we obtain the inequality $1+16a\ge a^2$. Solving this inequality gives $x\in \{0,\,1,\,\cdots,\,16\}$. In addition, $1+16a$ is a perfect square, we get $x\in \{0,\,3,\,5,\,14\}$. Only $x=0,\,5$ give integer value of $b$.

The solutions are hence $(0,\,1)$, $(0,\,-1)$, $(5,\,4)$ and $(5,\,-4)$.