## Task $2 / 75$

Determine all non-negative integers $m$ and $n$ that satisfy the equation $3 \cdot 2^{m}+1=n^{2}$.
From the given equation, by equivalent transformation, we get

$$
3 \cdot 2^{m}=n^{2}-1=(n+1)(n-1)
$$

Since the left side of the equation is even (if $m \neq 0$), both $n+1$ and $n-1$ must be even; thus, the left side of the equation must be factored into a product of two even factors. We now distinguish two cases:

Case 1: $n+1=3 \cdot 2^{x} ; n-1=2^{y}$

Case 2: $n+1=2^{x} ; n-1=3 \cdot 2^{y}$

In both cases, $x+y=m$ holds. From the second equations, we get $n=2^{y}+1$ and $n=3 \cdot 2^{y}+1$ respectively. Substituting these into the first equations, we obtain

Case 1: $2^{y}+2=3 \cdot 2^{x} ; 2^{y-1}+1=3 \cdot 2^{x-1}$

Case 2: $3 \cdot 2^{y}+2=2^{x} ; 3 \cdot 2^{y-1}+1=2^{x-1}-1$

In each of these equations, one side is odd; consequently, the other side must also be odd. This is only the case if $2^{x-1}=1$ (Case 1) or $2^{y-1}=1$ (Case 2). This leads to

Case 1: $x=1 ; y=2$; Case 2: $y=1 ; x=3$. Thus, we get $m=3 ; n=5$ and $m=4 ; n=7$.

It remains to examine the case $m=0$ that was initially excluded. For this case, the given equation yields the third solution $m=0 ; n=2$.