\( n \equiv 1 \pmod{3}, n \equiv 1 \pmod{5} \Rightarrow n \equiv 1 \pmod{15} \)

February 22, 2026

$n \equiv 1 \pmod{3}, n \equiv 1 \pmod{5} \Rightarrow n \equiv 1 \pmod{15}$

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\( n \equiv 4 \pmod{5} \Rightarrow n^2 \equiv 16 \equiv 1 \pmod{5} \).
\( n \equiv 1 \text{ or } 2 \pmod{3} \)
\( n \equiv 1 \text{ or } 4 \pmod{5} \)
\( n \equiv 1 \pmod{3}, n \equiv 4 \pmod{5} \): Solve \( n = 5k + 4 \). Then \( 5k + 4 \equiv 1 \pmod{3} \Rightarrow 2k + 1 \equiv 1 \Rightarrow 2k \equiv 0 \Rightarrow k \equiv 0 \pmod{3} \Rightarrow k = 3m \Rightarrow n = 15m + 4 \). So smallest is \( n = 4 \).
\( n \equiv 2 \pmod{3}, n \equiv 1 \pmod{5} \): Let \( n = 5k + 1 \). Then \( 5k + 1 \equiv 2 \pmod{3} \Rightarrow 2k + 1 \equiv 2 \Rightarrow 2k \equiv 1 \Rightarrow k \equiv 2 \pmod{3} \Rightarrow k = 3m + 2 \Rightarrow n = 5(3m+2)+1 = 15m + 11 \). Smallest is \( n = 11 \).
\( n \equiv 2 \pmod{3}, n \equiv 4 \pmod{5} \): \( n = 5k + 4 \). Then \( 5k + 4 \equiv 2 \pmod{3} \Rightarrow 2k + 1 \equiv 2 \Rightarrow 2k \equiv 1 \Rightarrow k \equiv 2 \pmod{3} \Rightarrow k = 3m + 2 \Rightarrow n = 15m + 14 \). Smallest \( n = 14 \).

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