## Task A-3.7.

Determine all pairs $(a, b)$ of natural numbers for which

$$
a^{b}-b^{a}+2^{a}=17 a^{4}-2 b^{2}+52
$$
## Solution.

Considering the parity of the expression in the equation, it follows that $b$ must be an even number.

1 point

First, assume that $a$ is an even number. Let's look at the equation modulo 8. The right side gives a remainder of 4 when divided by 8. On the other hand, if both numbers $a$ and $b$ were greater than or equal to 4, then the left side would be divisible by 8, which leads to a contradiction. Therefore, $a \leqslant 3$ or $b \leqslant 3$.

Since $a$ and $b$ are even, we need to check the cases $a=2$ and $b=2$.

1) If $a=2$, the equation reduces to:

$$
\begin{aligned}
2^{b}-b^{2}+4 & =17 \cdot 16-2 b^{2}+52 \\
2^{b}+b^{2} & =320
\end{aligned}
$$

For $b=8$, we get equality and one solution $(a, b)=(2,8)$.

1 point

The left side of the equation increases as $b$ increases, so for all $b$ different from 8, the equality does not hold, and we have no more solutions.

2) If $b=2$, the equation reduces to:

$$
\begin{aligned}
a^{2}-2^{a}+2^{a} & =17 a^{4}-2 \cdot 4+52 \\
17 a^{4}-a^{2}+44 & =0
\end{aligned}
$$

Since for all natural numbers $a$, $a^{4} \geqslant a^{2}$, the left side of the equation is clearly strictly greater than zero, so this equation has no natural solutions.

Now assume that $a$ is an odd number. Again, look at the equation modulo 8. Since the square of an odd number gives a remainder of 1 when divided by 8, the right side gives a remainder of 5. On the other hand, if $a$ were greater than or equal to 3, then for the same reason, the left side would give a remainder of 1 when divided by 8, which again leads to a contradiction. Therefore, $a \leqslant 2$.

Since $a$ is odd, the only possibility is $a=1$. The equation then reduces to:

$$
\begin{aligned}
1-b+2 & =17-2 b^{2}+52 \\
2 b^{2}-b-66 & =0 \\
(2 b+11)(b-6) & =0
\end{aligned}
$$

which gives us the solution $(a, b)=(1,6)$.

Finally, the pairs $(a, b)$ we are looking for are $(1,6)$ and $(2,8)$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

4th grade - high school - A variant

## February 17, 2021.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.