Solving the Diophantine Equation $a^4 + 79 + b^4 = 48ab$

[anemone]

Find all integer solutions $(a,\,b)$ satisfying $a^4+79+b^4=48ab$.

 

[kaliprasad]

Find all integer solutions $(a,\,b)$ satisfying $a^4+79+b^4=48ab$.

Spoiler

we make following observations

No 1) both are positive / -ve
No 2) one is odd and another even as RHS is even
No 3) The lower number is $< 5$ as $2 * 5^4 >= 1200$
No 4) x mod 3 cannot be zero nor y mod 3 as 79 mod 3 = 1 and RHS mod 3 = 0
No 5) if $(x,y) $ is a solution then $(-x,-y)$ is a solution
based on this we need to check (1,2), (1,4) so on
(2,5) so on
(4,5) so on

using above 3 conditions we get (1,2) a solution

(1,4) gives LHS = 336 > RHS
(2,5) gives LHS larger
(4,5) give LHS = RHS

so solution set = $(1,2).(2,1),(4,5),(5,4),(-1,-2),(-2,-1),(-4,-5),(-5,-4)$

 

[anemone]

Spoiler

we make following observations

No 1) both are positive / -ve
No 2) one is odd and another even as RHS is even
No 3) The lower number is $< 5$ as $2 * 5^4 >= 1200$
No 4) x mod 3 cannot be zero nor y mod 3 as 79 mod 3 = 1 and RHS mod 3 = 0
No 5) if $(x,y) $ is a solution then $(-x,-y)$ is a solution
based on this we need to check (1,2), (1,4) so on
(2,5) so on
(4,5) so on

using above 3 conditions we get (1,2) a solution

(1,4) gives LHS = 336 > RHS
(2,5) gives LHS larger
(4,5) give LHS = RHS

so solution set = $(1,2).(2,1),(4,5),(5,4),(-1,-2),(-2,-1),(-4,-5),(-5,-4)$

Well done, kaliprasad! And thanks for participating!:)

 

[kaliprasad]

Well done, kaliprasad! And thanks for participating!:)

Thanks for the same. I would like to see your solution Anemone

 

[anemone]

Thanks for the same. I would like to see your solution Anemone

Fair point!:)

But I didn't solve it myself, having said so, one has to have the complete solution for his/her posted challenge problem. So, I do have the solution that I wanted to share with you and all here:

Spoiler

If $(a,\,b)$ is a solution, so are $(b,\,a)$, $(-a,\,-b)$ and $(-b,\,-a)$.

Also, $ab>0$, so we must have $a,\,b$ both positive or both negative.

Suppose that $(a,\,b)$ is a solution, with $a\ge b>0$, we see that

$a^4+(b^4+79)=48ab$

$\dfrac{a^4}{a}+\dfrac{(b^4+79)}{a}=\dfrac{48ab}{a}$

$a^3+\dfrac{(b^4+79)}{a}=48b$

$a^3+\dfrac{(b^4+79)}{a}\le 48a$

It follows that $a^3\le 48a$ or more simply $a^2\le 48$ and thus $|a|$ and $|b|$ are bounded by 6. Also, one must be even and the other odd. It follows quickly that the only solution are:

$(a,\,b)=(-4,\,-5),\,(-5,\,-4),\,(-1,\,-2),\,(-2,\,-1),\,(1,\,2),\,(2,\,1),\,(4,\,5),\,(5,\,4)$