Algebra Problem #1

Question

How can this be solved in a simple way?

Find all positive integer pairs $(a,b)$ such that $$a^3 - b^3 - 6b^2 + 6a^2 = 1863$$

I'm not exactly sure how this can be solved simply. I've tried factoring, but I can't get much farther.

Answer 1

So you are interested in solving the following Diophantine equation in the positive integers: $$ a^3 - b^3 + 6(a^2 - b^2) = (a-b)\left[a^2+ab+b^2+6a+6b\right] = 1863 $$ If $a=b$, then there are no solutions.

If $a>b> 0$, then there are only finitely many cases to check, since the following holds when $b\geq 25$: $$ \left|(a-b)\left[a^2+ab+b^2+6a+6b\right]\right|>3b^2>1863 $$

If $b>a>0$, then there are only finitely many cases to check since the following holds when $a\geq 25$: $$ \left|(a-b)\left[a^2+ab+b^2+6a+6b\right]\right|>3a^2>1863 $$ I have just completed the script to exhaust over these possibilities, and it would appear that the only solution is $(a,b)=(14,11)$.

Answer 2

As Valborg wrote, $a^3 - b^3 + 6(a^2 - b^2) = (a-b)(a^2+ab+b^2+6a+6b) = 1863 =9\cdot 207 =81\cdot 23 =3^4\cdot 23 $.

Let $a-b = c$, so $a = b+c$. Then $a^2+ab+b^2+6a+6b =(b+c)^2+(b+c)b+b^2+6(b+c)+6b =3 b^2 + 3 b c + 12 b + c^2 + 6 c $ so $c(3 b^2 + 3 b c + 12 b + c^2 + 6 c) =3^4 23 $.

If $3 \not\mid c$ then $3 \not \mid (3 b^2 + 3 b c + 12 b + c^2 + 6 c) $ which can;t be. So, let $c = 3d$,

Then $3^4 23 =3d(3 b^2 + 9 b d + 12 b + 9d^2 + 18 d) =9d( b^2 + 3 b d + 4 b + 3d^2 + 6 d) $ so $3^2 23 =207 =d( b^2 + 3 b d + 4 b + 3d^2 + 6 d) =d((b+d)^2 + b d + 4 b + 2d^2 + 6 d) $.

If $23 | d$, the right side is too big. Therefore $d = 1, 3,$ or $9$.

If $d=1$ then $207 = b^2 + 3 b + 4 b + 3 + 6 =b^2+7b+9 $ so $b=-18, 11$. So $b=11, c=3, a=14 $.

If $d=3$ then $69 = b^2 + 9 b + 4 b + 27 + 18 = b^2 + 13 b +45 $ which has no integer roots.

If $d = 9$ then $23 = b^2 + 27 b + 4 b + 243 + 54 = b^2 + 31 b + 297 $ which has no integer roots.

Therefore $a=14, b=11$ is the only solution.

Answer 3

Noodling around

$a^3 - b^3 - 6b^2 + 6a^2 = 1863$

$a^3 + 6a^2 + 12a + 8 -b^3-6b^2 -12b - 8 = 1863 +12(a-b)$

$(a + 2)^3 - (b + 2)^3 = 1863 + 12(a-b)$

$(a - b)((a+2)^2 + (a+2)(b+2) + (b+2)^2 = 1863 + 12(a-b)$

Assuming $a \ne b$ (in which case $0=1863$ so we know that cant be) then

$(a + 2)^2 +(a+2)(b+2) + (b+2)^2 = \frac {1863}{a-b} + 12$

So $a-b|1863 = 3^4*23$

As $a$ and $b$ are both positive, we know that $a > b$

Let $a = b + k$ and so

$(a +2)^2 + (a+2)(a +2 - k) + (a+ 2 - k)^2 = \frac {1863}{k} + 12$

Which can be solved by the quadratic formula.

$(a^2 + 4a + 4) + (a^2 + 4a + 4) - (a+2)k + (a^2 + 4a + 4) -2k(a+2) + k^2 =\frac {1863}{k} + 12$