## Task A-3.3.

Determine all ordered triples of natural numbers $(a, b, c)$ for which $2^{a}+2^{b}+2^{c}+3$ is a square of some natural number.
## Solution.

Since the expression $2^{a}+2^{b}+2^{c}+3$ is symmetric in $a, b$, and $c$, we can assume without loss of generality that $a \leqslant b \leqslant c$.

If $a \geqslant 2$, then $2^{a}+2^{b}+2^{c}+3$ gives a remainder of 3 when divided by 4 and therefore cannot be a perfect square. Thus, it must be that $a=1$.

Furthermore, if $b \geqslant 3$, then $2^{a}+2^{b}+2^{c}+3=2^{b}+2^{c}+5$ gives a remainder of 5 when divided by 8, and therefore cannot be a perfect square.

It remains to check the cases when $b=1$ and $b=2$.

If $b=1$, then $2^{a}+2^{b}+2^{c}+3=2^{c}+7$ must be a perfect square.

If $c \geqslant 2$, then $2^{c}+7$ will give a remainder of 3 when divided by 4, and therefore $2^{c}+7$ cannot be a perfect square.

By direct verification, we see that for $c=1$ we get a solution, since $2^{1}+2^{1}+2^{1}+3=9$ is a perfect square.

If $b=2$, then $2^{a}+2^{b}+2^{c}+3=2^{c}+9$ must be a perfect square, i.e., $2^{c}+9=k^{2}$ for some natural number $k$. Using the difference of squares, we can write the equation as

$$
2^{c}=k^{2}-9=(k-3)(k+3)
$$

From the above equality, it follows that $k-3$ and $k+3$ are powers of 2, i.e., $k-3=2^{c_{1}}$ and $k+3=2^{c_{2}}$ where $c_{1}$ and $c_{2}$ are non-negative integers such that $c=c_{1}+c_{2}$ and $c_{1}<c_{2}$. Subtracting these equalities, we get

$$
6=2^{c_{2}}-2^{c_{1}}=2^{c_{1}}\left(2^{c_{2}-c_{1}}-1\right)
$$

from which we conclude that $c_{1}=1$ and $c_{2}=3$, i.e., $c=4$. In this case, $k=5$.

Thus, under the assumption $a \leqslant b \leqslant c$, the solutions are $(a, b, c)=(1,1,1),(1,2,4)$.

All possible triples of numbers $(a, b, c)$ are obtained by permuting the above solutions:

$$
(a, b, c)=(1,1,1),(1,2,4),(1,4,2),(2,1,4),(2,4,1),(4,1,2),(4,2,1). \quad 1 \text { point }
$$