From Fermat's Little theorem, for primes $p$, we know that

$$ p \mid 2^p - 2. $$

For which primes $p$ does

$$p^2 \mid 2^{p} - 2 ?$$

Most people when first seeing this question, would try small cases of $p$, and realize that it doesn't work. They may then look at

$$\frac{(1+1) ^p - 2}{p} = \frac{{ p \choose 1} + { p\choose 2} + \ldots + { p \choose p-1}}{p} \equiv \frac{1}{1} + \frac{-1}{2} + \frac{1}{3} + \ldots + \frac{(-1)^{p}}{p-1} \pmod{p}$$

and try and prove that it is not 0.

As it turns out, the Wieferich primes satisfy $p^2 | 2^p-2$. There are only 2 known examples of such primes, namely $p=1093, 3511$. There are no other examples less that $10^{17}$.

From Fermat's Little theorem, for primes $p$, we know that

$$ p \mid 2^p - 2. $$

For which primes $p$ does

$$p^2 \mid 2^{p} - 2 ?$$

Most people when first seeing this question, would try small cases of $p$, and realize that it doesn't work. They may then look at

$$\frac{(1+1) ^p - 2}{p} = \frac{{ p \choose 1} + { p\choose 2} + \ldots + { p \choose p-1}}{p} \equiv \frac{1}{1} + \frac{-1}{2} + \frac{1}{3} + \ldots + \frac{(-1)^{p}}{p-1} \pmod{p}$$

and try and prove that it is not 0.

As it turns out, the Wieferich primes satisfy $p^2 | 2^p-2$. There are only 2 known examples of such primes, namely $p=1093, 3511$. There are no other examples less than $10^{17}$.

From Fermat's Little theorem, for primes $p$, we know that

$$ p \mid 2^p - 2. $$

For which primes $p$ does

$$p^2 \mid 2^{p} - 2 ?$$

Most people when first seeing this question, would try small cases of $p$, and realize that it doesn't work. They may then look at

$$\frac{(1+1) ^p - 2}{p} = \frac{{ p \choose 1} + { p\choose 2} + \ldots + { p \choose p-1}}{p} \equiv \frac{1}{1} + \frac{-1}{2} + \frac{1}{3} + \ldots + \frac{(-1)^{p}}{p-1} \pmod{p}$$

and try and prove that it is not 0.

As it turns out, the Wieferich primes satisfy $p^2 | 2^p-2$. There are only 2 known examples of such primes, namely $p=1093, 5311$. There are no other examples less that $10^{17}$.

From Fermat's Little theorem, for primes $p$, we know that

$$ p \mid 2^p - 2. $$

For which primes $p$ does

$$p^2 \mid 2^{p} - 2 ?$$

Most people when first seeing this question, would try small cases of $p$, and realize that it doesn't work. They may then look at

$$\frac{(1+1) ^p - 2}{p} = \frac{{ p \choose 1} + { p\choose 2} + \ldots + { p \choose p-1}}{p} \equiv \frac{1}{1} + \frac{-1}{2} + \frac{1}{3} + \ldots + \frac{(-1)^{p}}{p-1} \pmod{p}$$

and try and prove that it is not 0.

As it turns out, the Wieferich primes satisfy $p^2 | 2^p-2$. There are only 2 known examples of such primes, namely $p=1093, 3511$. There are no other examples less that $10^{17}$.