Which positive integers $a$ and $b$ make $(ab)^2-4(a+b) $ a square of an integer?

Question

Which positive integers $a$ and $b$ make $(ab)^2-4(a+b) $ a square of an integer?

I saw this in quora, and found that the only solutions with $a \ge b > 0$ are $(a, b, (ab)^2–4(a+b)) = (5, 1, 1)$ and $(3, 2, 16)$. Another “solution” is $a = b = 2, (ab)^2–4(a+b) = 0$.

My solution is messy and computational, and I wonder if there is a more elegant solution.

Here is my solution.

Assume $a \ge b$ and write $n^2 = (ab)^2–4(a+b)$ so $n < ab$.

Let $n = ab-k$ where $ab > k>0$ so $(ab)^2–4(a+b) = (ab-k)^2 = (ab)^2–2kab+k^2$ or $k^2–2kab+4(a+b) = 0$.

Then

$\begin{array}\\ k &= \dfrac{2ab-\sqrt{4a^2b^2–16(a+b)}}{2} \qquad \text{(use "-" since } k < ab)\\ &= ab-\sqrt{a^2b^2–4(a+b)}\\ &=(ab-\sqrt{a^2b^2–4(a+b)})\dfrac{ab+\sqrt{a^2b^2–4(a+b)}}{ab+\sqrt{a^2b^2–4(a+b)}}\\ &=\dfrac{4(a+b)}{ab+\sqrt{a^2b^2–4(a+b)}}\\ \end{array} $

Therefore, since $k \ge 1, 4(a+b) \ge ab$ so $0 \ge ab-4(a+b) = ab-4(a+b)+16–16 =(a-4)(b-4)-16$ or $16 \ge (a-4)(b-4)$. This gives a finite number of possible $a, b$, all at least $4$.

Computation shows that none of these are solutions.

To get the possible values of $a$ and $n$ in terms of $b$ for any fixed $b$, do this:

Since $n^2 = a^2b^2-4(a+b)$,

$\begin{array}\\ b^2n^2 &= a^2b^4-4b^2a-4b^3\\ &= a^2b^4-4b^2a+4-4b^3–4\\ &=(b^2a-2)^2–4(b^3+1)\\ \end{array} $

so $4(b^3+1) = (b^2a-2)^2-b^2n^2 = (b^2a-2-bn)(b^2a+bn) $.

For each factorization $r*s = 4(b^3+1)$, try $r=b^2a-2-bn, s=b^2a-2+bn$.

This gives $s-r=2bn$, so if $2b$ divides $s-r$, then $n=\dfrac{s-r}{2b}$. Adding $s$ and $r$, $2b^2a-4=s+r$ so if $2b^2$ divides $s+r+4$, then $a = \dfrac{s+r+4}{2b^2}$.

This allows us to compute all solutions for any fixed value of b. Running this for $1 \le b \le 16$ gives the solutions above.

For $a \ge b \ge 5$, the restriction $16 \ge= (a-4)(b-4)$ gives a finite set of possibilities which computation shows yields no additional solutions.

I sure would like to see a more elegant solution. Also, this messy algebra provides lots of opportunities for errors.

Answer 1

We can find an upper bound for their "product" in the following way: \begin{align} {\left( {ab} \right)^2} - 4\left( {a + b} \right) &< {\left( {ab} \right)^2} \ \ (\because a,b>0) \\ {\left( {ab} \right)^2} - 4\left( {a + b} \right) &\le {\left( {ab - 1} \right)^2} \\ (2ab-1)-4(a+b) &\le 0\\ ab - 2(a+b) - \frac 12 \color{blue}{+4} &\le 0 \color{blue}{+4}\\ \left( {a - 2} \right)\left( {b - 2} \right) &\le \frac{9}{2} \\ \left( {a - 2} \right)\left( {b - 2} \right) &\le 4 \\ \end{align}

Now, seeing some cases should finish the work.

Answer 2

If $(ab)^2-4(a+b)$ is greater than $(ab-1)^2$ then it cannot be a square, since it is strictly between two consecutive squares. Hence

$$(ab)^2-4(a+b) \le (ab-1)^2=(ab)^2-2ab+1$$ $$2ab-4a-4b-1\le0$$ $$2(a-2)(b-2)=2ab-4a-4b+8\le 9$$

which, again, gives a finite set of possibilities to be checked.

WLOG suppose $a \ge b$. We need only consider the cases:

1. $b=1,2$
2. $b=3, a\le6$
3. $b\ge 4, a < 3$ (this case contradicts $a\ge b$)

For $b=1$, $(ab)^2-4(a+b) = a^2-4a-4$. For $a\ge7$, $a^2-4a-4> a^2-6a+9=(a-3)^2$. But $a^2 -4a-4 < a^2-4a+4=(a+2)^2$. So we only need to check $1 \le a \le 6$.

For $b=2$, $(ab)^2-4(a+b) = 4a^2-4a-8$, which is a square only if $a^2-a-2$ is. For $a\ge4$, $a^2-a-2> a^2-2a+1 =(a-1)^2$. But $a^2-a-2<a^2$, so we only need to check $a=2,3$.

For $b=3$, $(ab)^2-4(a+b) = 9a^2-4a-12$, and we only need to check $3 \le a \le 6$.