Finding solutions in $\mathbb{Z}_{+}$

Question

Find all the triplets of positive integers $(a, b, c)$ such that: $a^2+b+3=(b^2-c^2)^2$

Answer 1

Note that:

$$\text{L.H.S}>0\rightarrow \text{R.H.S}>0\rightarrow b\ne c\rightarrow (b-c)^2\ge 1$$

$$\Rightarrow a^2+b+3=(b-c)^2(b+c)^2\ge (b+c)^2\ge (b+1)^2 (\text{c}\in N^+) $$

$$\Leftrightarrow a^2+b+3\ge b^2+2b+1\Leftrightarrow a^2\ge b^2+b-2$$

1. $b>2\rightarrow a^2>b^2\rightarrow a>b>2$

$\Rightarrow a^2<a^2+b+3<(a+1)^2$ which has $a^2+b+3$ isnt square number.

But $a^2+b+3=(b^2-c^2)^2$ is square.

It is obviously wrong

2. $b=1$ we have: $$\Rightarrow (c^2-1+a)(c^2-1-a)=4$$

It isnt difficult to see that there isnt root.

3. $b=2$ we have: $$\Rightarrow (4-c^2-a)(4-c^2+a)=5$$

$\rightarrow a=2;c=1$

conclude: $(a;b;c)=(2;2;1)$ is the triplet of positive integers such that $a^2+b+3=(b^2-c^2)^2$