## Subject II

Calculate the values of the natural numbers $\mathrm{a}, \mathrm{b}$, and $\mathrm{p}$, where $\mathrm{p}$ is a prime number, knowing that $a^{2}+a=p^{2^{b}}+2$.

Mathematical Gazette, E:14569
Solution. The equality from the statement is equivalent to $a(a+1)=p^{2^{b}}+2(1)$, but $a(a+1)$ is an even number, so $p^{2^{b}}$ is an even number and since $p$ is a prime number, we deduce that $p=2$; substituting in relation (1) and factoring out 2: $a(a+1)=2\left(2^{2^{b}-1}+1\right)(2)$; if $a=1 \Rightarrow 2=2\left(2^{2^{b}-1}+1\right) \Leftrightarrow 2^{2^{b}-1}+1=1 \Leftrightarrow 2^{2^{b}-1}=0$ impossible (3); the numbers $a$ and $a+1$ being consecutive are coprime (4); the numbers 2 and $2^{2^{b}-1}+1$ are coprime (5); from (3), (4) and (5) it follows that $\mathrm{a}=2$ is the only possible situation and substituting in (1) we get $2^{2^{b}}+2=6$, $\Leftrightarrow$ $2^{b}=2^{1} \Leftrightarrow b=1$.

## Grading Rubric:

```
- writes (1)
    (1p)

```

![](https://cdn.mathpix.com/cropped/2024_06_07_1933c58db8eb232513deg-2.jpg?height=45&width=645&top_left_y=1685&top_left_x=186)

```
- analyzes \(\mathrm{a}=1\) and deduces (3) ..... (1p)
- observes (4) and (5) ................... (2p)
- \(\mathrm{a}=2\) is the only possibility ..... (1p)

```

![](https://cdn.mathpix.com/cropped/2024_06_07_1933c58db8eb232513deg-2.jpg?height=43&width=643&top_left_y=1869&top_left_x=187)

```
    Total \(7 \mathbf{p}\)
```