MHB Determine all pairs of integers

[anemone]

Determine all pairs of integers $(a, b)$ satisfying the equation $b(a+b)=a^3-7a^2+11a-3$.

 

[Opalg]

This is not a solution. (Emo)

[sp]The pairs $(a,b) = (1,-2),\ (1,1),\ (2,-1),\ (6,-9),\ (6,3)$ are solutions. I believe that these five are the only solutions but I do not see how to prove that.[/sp]

 

[mathbalarka]

(Emo) I believe anemone used some ineq here too, but that thing there seems suspiciously like an elliptic curve.

 

[Opalg]

(Emo) I believe anemone used some ineq here too, but that thing there seems suspiciously like an elliptic curve.

I can reduce the problem to finding solutions to the cubic diophantine equation $y^2 = x^3 - 67x - 66$. There are (at least) three solutions $x = -5, -1, 15$, but that's as far as I can go.

 

[mente oscura]

This is not a solution. (Emo)

[sp]The pairs $(a,b) = (1,-2),\ (1,1),\ (2,-1),\ (6,-9),\ (6,3)$ are solutions. I believe that these five are the only solutions but I do not see how to prove that.[/sp]

Hello.

Yes, I have come to the same conclusion: brute force

Spoiler

Restrictions:

1ª) a \ge{0}

2ª) If \ b=even \rightarrow{}4 \cancel{|}b

3ª) a \le {|b+3|}

4ª) b=\dfrac{-a \pm {} \sqrt{4a^3-27a^2+44a-12}}{2}

5ª) 4a^3-27a^2+44a-12=T^2

Pero, en fin: la fuerza bruta, referida.(Rofl)

(1,1), (1,-2), (2,-1), (6,3), (6, 9)

I do not know if there is more

Regards.

 

[anemone]

I want to apologize because I only respond to this thread days after. I am sorry...

Spoiler

I didn't solve this problem, in fact, I spent days to solve it but ended up with all futile attempts. Having said so, I do have a solution that is suggested by other in another math forum which I want to share it here:

We're asked to determine all pairs of integers $(a, b)$ satisfying the equation $b(a+b)=a^3-7a^2+11a-3$.

If we rewrite the equation, we get

$4b(a+b)=4(a^3-7a^2+11a-3)$

$4ab+4b^2=4a^3-28a^2+44a-12$

$4ab+4b^2+a^2=4a^3-27a^2+44a-12$

$(2b+a)^2=(a-2)(4a^2-19a+6)$

$(2b+a)^2=(a-2)(4(a-2)^2-3(a-2)-16)$

Let $x=a-2$ we then have

$(2b+a)^2=x(4x^2-3x-16)=x(x(4x-3)-16)$

We can tell that $x(4x^2-3x-16)$ is a perfect square, and hence

$x(4x^2-3x-16) \ge 0$

This gives us the solution sets of $-1\le x \le 0$ and $x \ge 3$.

For $-1\le x \le 0$:

$x=-1$ gives $a=1$, and $(2b+1)^2=9$ and $\therefore b=-2,1$.

$x=0$ gives $a=2$, and $(2b+2)^2=0$ and $\therefore b=-1$.

For $x \ge 3$, $gcd(x, x(x(4x-3)-16))|16$ and hence $(x, x(x(4x-3)-16))|16$.

For this case, we have

$x=4$ gives $a=6$, and $(2b+6)^2=144$ and $\therefore b=-9,3$.

and we're done.