I have read other proof, but I am stuck on the proof in my algebra class. Hope someone could help me. Thanks a lot.

Prove by contradiction. Let $ \{ p_1,\dots p_n\} $ be our finite primes with $p_i \equiv 2 (\text{mod3})$ $\forall i$.

Let $$m=1+p_1^{2}\dots p_n^{2}$$

Then $m\equiv 2 (\text{mod3})$.

By fundamental theorem of arithmetic $m=q_1\dots q_t$, where $q_i$ is prime $\forall i$

How can I prove that $q_i=3$ or $q_i\equiv1 (\text{mod3})$ $\forall I$ ?

Since if I prove it, then $m=q_1\dots q_t\equiv 1 \;\text{or}\; 3$ and we get the contradiction.

My professor have define $q_i'=p_1^2\dots p_i\dots p_n^2$, so that $m=q_i'p_i+1$. But I have no idea why we have to define this.


Since $m \equiv 2\;(\text{mod}\;3)$, we can't have any $q_i=3$ (else $m$ would be a multiple of $3$).

But if $q_i \equiv 1\;(\text{mod}\;3)$ for all $i$, then the product $q_1\cdots q_t$ would be congruent to $1$ mod $3$, contrary to $m \equiv 2\;(\text{mod}\;3)$.

It follows that $q_k \equiv 2\;(\text{mod}\;3)$ for some $k$.

But $m$ is not divisible by any of $p_1,...,p_n$, so $q_k$ is a new prime congruent to $2$ mod $3$.

That's the contradiction.


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