$a^2+a+3b$ and $b^2+b+3a$ both perfect square

Question

Find all posssible $(a,b)$ where $a$ and $b$ natural number (positive integer) such that $a^2+a+3b$ and $b^2+b+3a$ both perfect square

so, i try to assume that $a^2+a+3b=(a+m)^2$ with $m$ natural number then we have $a+3b=2ma+m^2$ similarly $b+3a=2nb+n^2$

any further manipulation make the equation be more difficult non linear diophantine that i cannot solve no matter how many times i tried. please help

Answer 1

You made a good start. Here we set aside the trivial solution $a=b=0$. All you need to do next is check through a few possibilities for $m$ and $n$; most potential values for $m$ and $n$ can be ruled out, because the two equations $$3b=m^2+(2m-1)a\quad\text{and}\quad 3a=n^2+(2n-1)b$$ would then lead to contradictions. First note that, if $m\geqslant2$, then $2m-1\geqslant3$ and so $b>a$ by the first equation; similarly, if $n\geqslant2$, then $a>b$ by the second equation. Therefore, we cannot have both $m\geqslant2$ and $n\geqslant2$. Note also that $m$ (or $n$) cannot equal $2$, because $2^2$ is not divisible by $3$. So, by symmetry, we may assume $m=1$; the other solutions can be recovered by swapping the roles of $a$ and $b$. Our first equation becomes $a=3b-1$, and then $2(5-n)b= n^2+3$. By parity, $n\neq4$. Also, it is easy to check that $n\neq1$. We are left with $n=3$. Consequently the only solutions are $$\{a,b\}=\{3,8\}.$$

Answer 2

You have the right general idea, but the following shows how to use perfect square lower and upper limits to determine the solution. First, due to symmetry, WLOG consider that $a \ge b$. Next, if natural numbers are meant to include $0$, then with $b = 0$, we have $a^2 + a$ and $3a$ must both be perfect squares. However, for $a \gt 0$, we have $a^2 \lt a^2 + a \lt a^2 + 2a + 1 = (a + 1)^2$, so there are no solutions. Thus, the only solution is $(0,0)$. Otherwise, for $b \ge 1$, we then have

$$a^2 \lt a^2 + a + 3b \lt a^2 + 4a + 4 \;\;\to\;\; a^2 \lt a^2 + a + 3b \lt (a+2)^2 \tag{1}\label{eq1A}$$

Since $a^2 + a + 3b$ is a perfect square, this means

$$a^2 + a + 3b = (a+1)^2 = a^2 + 2a + 1 \;\to\; 3b = a + 1 \;\to\; a = 3b - 1 \tag{2}\label{eq2A}$$

We then have from the second second perfect square expression that

$$b^2 + b + 3(3b - 1) = b^2 + 10b - 3 \tag{3}\label{eq3A}$$

We have it's $\gt b^2$, but it's also $\lt (b+5)^2$, so for some integer(s) $1 \le c \le 4$, we get that

$$b^2 + 10b - 3 = (b + c)^2 = b^2 + 2bc + c^2 \to b(10 - 2c) = c^2 + 3 \to b = \frac{c^2 + 3}{10 - 2c} \tag{4}\label{eq4A}$$

Due to $b$ being a positive integer, since the denominator is even, the numerator must be as well, i.e., $c$ is odd. Only $c = 3$ works to then get that $b = 3$. From \eqref{eq2A}, this then gives that $a = 8$. To confirm that this results in both expressions being perfect squares, we get $a^2 + a + 3b = 81 = 9^2$ and $b^2 + b + 3a = 36 = 6^2$. Thus, since $a$ and $b$ can be switched around, there is only the $(0,0)$ solution mentioned initially, and the $2$ additional solutions, to get

$$(a,b)\in\{(0,0),(8,3),(3,8)\} \tag{5}\label{eq5A}$$