Why 3^2463 ≡ 3 (mod 8)

For any fixed natural number m greater than 1, we say that the integer a is congruent to the integer b modulo m if a − b is divisible by m. Please find the remainder when 3^2463 is divided by 8.

To compute $3^{2463}\bmod8$, observe the pattern of powers of 3 modulo 8:

$$ \begin{aligned} 3^1 &\equiv 3 \pmod{8},\\ 3^2 &\equiv 9\equiv1 \pmod{8},\\ 3^3 &\equiv3^2\cdot3\equiv1\cdot3\equiv3\pmod{8},\\ 3^4 &\equiv(3^2)^2\equiv1^2\equiv1\pmod{8}, \end{aligned} $$

so the residues cycle with period 2:

$$ 3^{\text{odd}}\equiv3,\quad 3^{\text{even}}\equiv1\pmod8. $$

Since $2463$ is odd,

$$ 3^{2463}\equiv3\pmod8. $$

Answer: The remainder when $3^{2463}$ is divided by 8 is $\boxed{3}$.

Why do we expect that this pattern always holds?

The reason the pattern

$$ 3^1\equiv3,\quad 3^2\equiv1,\quad 3^3\equiv3,\quad3^4\equiv1,\;\dots $$

repeats every two exponents comes down to the single fact

$$ 3^2 = 9 = 8 + 1 \;\;\Longrightarrow\;\; 3^2 \equiv 1 \pmod8. $$

Once you know that $3^2\equiv1\pmod8$, it follows immediately that for any integer $k$,

• $$ 3^{2k}=(3^2)^k\equiv1^k\equiv1\pmod8, $$
• $$ 3^{2k+1}=3\cdot3^{2k}\equiv3\cdot1\equiv3\pmod8. $$

Thus the residues must alternate $3,1,3,1,\dots$ with period 2.

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A bit more structure: the multiplicative group of units modulo 8

• The units in $\mathbb{Z}/8\mathbb{Z}$ are those integers coprime to 8, namely $\{1,3,5,7\}$.

• This set forms a finite group under multiplication mod 8, whose order is $\varphi(8)=4$.

• By Lagrange’s theorem, the order (the size) of any element in a finite group must divide the size of the group. Here:

  $$ \text{order}(3)\;|\;4. $$

• We check directly that $3^2\equiv1\pmod8$, so the order of 3 is exactly 2, and hence its powers repeat with period 2.

Either way—whether by the elementary identity $3^2\equiv1\pmod8$ or by invoking the structure of $(\mathbb{Z}/8\mathbb{Z})^\times$—you see why the cycle length is 2, and why $3^{2463}\equiv3$.

I’m not sure why ChatGPT wrote “period 2” with a narrow no-break space (NNBSP) instead of a regular space (as in “period 2”).