1996 Irish Mathematical Olympiad

prime numbers defination

another relevance wiki , wiki of day . Why 2 is prime?

Let be $\text{p = 2}$

$\text({2}^{2})$ + $\text({3}^{2})$

$\text{(4+9 = 13)}$

$\text{13}^{1}$ = $\text{2}^{2}$ + $\text{3}^{2}$

Note : If $\text{(p = prime)}$ , the root of $\text{(n)}$ must be not perfect.

$\therefore{ always ~~ \text({n}^{1})}$ = ( $\text{2}^{prime}$ + $\text{3}^{prime}$ ) .

Let's first consider the case $p=2$ . Then $2^2 + 3^2 = 13$ . $13$ is not a perfect power, just $13 = 13^1$ . Let's now consider the bigger case of odd $p$ . If $p$ is odd then we have that $2^p + 3^p = (2 + 3)(2^{p-1} - 2^{p-2} 3 + ... + 3^{p-1}) = 5(2^{p-1} + 2^{p-2} 3 + ... + 3^{p-1})$ . The equality given in the problem would then imply that $a^n \equiv 0 \mod 5 \implies a \equiv 0 \mod 5$ . This means that we can say that $a = 5a_0$ and rephrase the equality in terms of this new variable:

$2^p + 3^p = 5^n a_0 ^n$

Dividing both sides by 5 we get:

$(2^{p-1} - 2^{p-2} 3 + ... + 3^{p-1}) = 5^{n-1} a_0 ^n$

If we assume that $n > 1$ then this implies that $2^{p-1} + 2^{p-2} 3 + ... + 3^{p-1} \equiv 0 \mod 5$ . We will prove that this is impossible:

Notice that $3 \equiv -2 \mod 5$ . This implies that $2^{p-1} - 2^{p-2} 3 + ... + 3^{p-1} = \sum_{n=0} ^{p-1} (-1)^n 2^{p-1-n} 3^n \equiv \sum_{n=0}^{p-1} (-1)^{2n} 2^{p-1} \equiv 2^{p-1} \sum_{n=0}^{p-1} 1 \equiv 2^{p-1} p \not \equiv 0 \mod 5$ .

The last congruence is true as long as $p \neq 5$ . As such, we must check that case too: $2^5 + 3^5 = 275 = 5^2 \times 11$ which is also not a perfect power.

This contradiction came from the assumption that $n>1$ . Therefore, it must be that $n=1$ always.