Answer
See explanation
Work Step by Step
Let
\[
N = p_1 \cdot p_2 \cdots p_n + 1,
\]
with \(p_1=2\) and \(p_2, \dots, p_n\) being odd primes (since every prime besides 2 is odd). Write
\[
N = 2\cdot (p_2p_3 \cdots p_n) + 1.
\]
Denote
\[
R = p_2p_3\cdots p_n.
\]
Because each \(p_i\) for \(i\ge2\) is odd, their product \(R\) is odd. Every odd integer is congruent to either 1 or 3 modulo 4. In either case, we have:
- If \(R \equiv 1 \pmod{4}\), then
\[
2R \equiv 2\cdot 1 = 2 \pmod{4}\quad \text{and}\quad N = 2R+1 \equiv 2+1 = 3 \pmod{4}.
\]
- If \(R \equiv 3 \pmod{4}\), then
\[
2R \equiv 2\cdot 3 = 6 \equiv 2 \pmod{4}\quad \text{and}\quad N = 2R+1 \equiv 2+1 = 3 \pmod{4}.
\]
Thus, in either case,
\[
N \equiv 3 \pmod{4}.
\]
This means that \(N\) can be written in the form
\[
N = 4k + 3,
\]
for some integer \(k\).
This completes the proof.
