To determine whether $2008$ is a quadratic residue modulo $257$, we can use properties of modular arithmetic and the Law of Quadratic Reciprocity. Here’s the process:

Step 1: Simplify the Problem Using Modular Reduction

Since we want to determine if $2008$ is a quadratic residue modulo $257$, we can reduce $2008$ modulo $257$: $2008 \mod 257 = 2008 - 257 \times 7 = 2008 - 1799 = 209$ Thus, the problem reduces to determining if $209$ is a quadratic residue modulo $257$.

Step 2: Apply the Legendre Symbol

The Legendre symbol $\left( \frac{209}{257} \right)$ will tell us if $209$ is a quadratic residue modulo $257$. Specifically:

• If $\left( \frac{209}{257} \right) = 1$, then $209$ (and thus $2008$) is a quadratic residue mod $257$.
• If $\left( \frac{209}{257} \right) = -1$, then $209$ (and thus $2008$) is a non-residue mod $257$.

Step 3: Use Quadratic Reciprocity

Since both $209$ and $257$ are odd primes, we can use the Law of Quadratic Reciprocity: $\left( \frac{209}{257} \right) = \left( \frac{257}{209} \right) \times (-1)^{\frac{209 - 1}{2} \cdot \frac{257 - 1}{2}}$ Calculating $\frac{209 - 1}{2} = 104$ and $\frac{257 - 1}{2} = 128$, both of which are even, the exponent $104 \cdot 128$ is even, so: $(-1)^{104 \cdot 128} = 1$ Thus: $\left( \frac{209}{257} \right) = \left( \frac{257}{209} \right)$

Step 4: Simplify $\frac{257}{209}$ Using Modular Reduction

Now, reduce $257$ modulo $209$: $257 \mod 209 = 257 - 209 = 48$ So we have reduced the expression to $\left( \frac{48}{209} \right)$.

Step 5: Further Reduce $\left( \frac{48}{209} \right)$

Since $48 = 16 \cdot 3$, we can split this using the property of the Legendre symbol: $\left( \frac{48}{209} \right) = \left( \frac{16}{209} \right) \cdot \left( \frac{3}{209} \right)$ Now:

1. $16$ is a perfect square ($16 = 4^2$), so $\left( \frac{16}{209} \right) = 1$.
2. We need to evaluate $\left( \frac{3}{209} \right)$.

Step 6: Apply Quadratic Reciprocity Again for $\left( \frac{3}{209} \right)$

Since both $3$ and $209$ are odd primes, we use reciprocity: $\left( \frac{3}{209} \right) = \left( \frac{209}{3} \right) \times (-1)^{\frac{3 - 1}{2} \cdot \frac{209 - 1}{2}}$ Calculating the exponent $1 \cdot 104 = 104$, which is even, we get: $(-1)^{104} = 1$ Thus: $\left( \frac{3}{209} \right) = \left( \frac{209}{3} \right)$ Reducing $209$ modulo $3$: $209 \mod 3 = 2$ So $\left( \frac{209}{3} \right) = \left( \frac{2}{3} \right) = -1$.

Conclusion

Since $\left( \frac{209}{257} \right) = -1$, $209$ (and therefore $2008$) is a quadratic non-residue modulo $257$.

Would you like more details on any step?

Here are some questions to expand on this problem:

1. What is the Law of Quadratic Reciprocity?
2. How do Legendre symbols help determine quadratic residues?
3. What are the properties of modular reduction?
4. Can quadratic reciprocity be used for composite numbers?
5. What is the significance of the exponent parity in quadratic reciprocity?

Tip: Always reduce large numbers modulo your target number early to simplify computations.