# Finding the integer solutions

Find all integer solutions of $$(a + b^2)(a^2 + b) = (a − b)^3.$$

Obviously $b = 0$ is one. But how to get other solutions?

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I would suggest starting by expanding both sides, combining both sides so that they are equal to zero. Then dividing by the obvious answer (b) –  kaine Jan 18 '13 at 17:53

On simplification, $$2b^3+(a^2-3a)b^2+b(a+3a^2)=0$$

If $$b\ne 0, 2b^2+(a^2-3a)b+(a+3a^2)=0$$ which is a quadratic equation in $b$

As $b$ is integer, the discriminant must be perfect square.

So, $$(a^2-3a)^2-4\cdot2\cdot(a+3a^2)=a^4-6a^3-15a^2-8a=f(a)$$(say)

Clearly, $f(0)=0=f(-1)$ so, $a^4-6a^3-15a^2-8a=a(a+1)(a^2-7a-8)=a(a+1)^2(a-8)$

So, we need $a(a-8)$ to be perfect square $\implies a(a-8)\ge 0\implies a\le0$ or $a\ge 8$.

Now, $$a^2-8a=(a-4)^2-16\iff 4^2=(a-4)^2-(a^2-8a)$$

Observe that

$n^2-(n-1)^2=2n-1$ is odd $\ne 16$

$n^2-(n-2)^2\le 16 \iff n \le5$

$n^2-(n-3)^2=6n-9$ is odd

$n^2-(n-4)^2=8n-16\le 16\iff n\le \frac{25}8$

$4^2=5^2-3^2,4^2-0^2$

So, we need

$(i)a^2-8a=3^2$ or $(a-4)^2=5^2$ both imply $a=-1,9$

or $(ii)a^2-8a=0\implies a=0,8$

There is no more solutions in integers.

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Here is one approach.

Expand both sides and equate terms, so we have:

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

$$a^{3} + a^{2}b^{2} + ab + b^{3} = a^{3} - 3a^{2}b + 3a b^2 - b^{3}$$

We can eliminate the $a^{3}$ terms, so we are left with:

$$2b^{3} + a^{2}b^{2} - 3ab^{2} + 3a^{2}b +ab = 0$$

Now, we have a Cubic in the form of:

$f(x) = jx^{3} + kx^{2} +lx + m = 0$, with $j = 2, k = a^{2}-3a, l = 3a^{2}+a$ and $m = 0$.

Using the solution for the cubic formula, we can arrive at the three roots:

$b = 0$

$b = (\frac{1}{4})(-a^{2} + 3a - \sqrt{a - 8}(a+1)\sqrt{a})$

$b = (\frac{1}{4})(-a^{2} + 3a + \sqrt{a - 8}(a+1)\sqrt{a})$

Now, you only want integer solutions, so, we see, by inspection, that at:

$a = -1, a = 0, a = 8, a = 9$, we get integers in $b$.

For $a = -1$, we have $b = -1$

For $a = 8$, we have $b = -10$

For $a = 0$, and $a = 9$, we have $b = 0$

Also note, for $b = 0$, the original equation reduces to, $a^{3} = a^{3}$, so you can also freely choose $a's$ (so you gets lots of integer solutions).

If you do an Implicit Plot using WA, you can see the $b = 0$ cases.

Regards

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Nice work... $+ 1^{+^+}$ –  amWhy May 8 '13 at 2:04
It's well-deserved ;-) –  amWhy May 8 '13 at 2:16

Suppose a = b+p p prime.Then a) a+b^2=p^2 (1) and a^2+b=p (2) simultanously. (2) leads to the equation 8*p +1 = d^2 whith only 1 solution for primes < 10^9; p=3 and (a,b)=(-3,-6);(2,-1) which are not solutions of (1) --->no solution. b) a+b^2=p (3) and a^2+b=p^2 (4) simultanously leads to the infinitely many solutions (a,b) = (p,0).

Unfortunately when we take a-b = p*q the product of two primes; this method is useless because there are far too many solutions to 8*p*q +1 = d^2.

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