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Exercises 16.8 Exercises

1.

Fill in all the details of Example 16.0.2 for the congruences x^2+5x+5\equiv 0\text{ (mod }5) and x^2+5x+7\equiv 0\text{ (mod }n)\text{.}

2.

Prove that if e>1\text{,} then there is no solution to

\begin{equation*} x^2\equiv -1\text{ (mod }2^e)\text{.} \end{equation*}

Use our knowledge of squares modulo 4.

3.

For what n does -1 have a square root modulo n\text{?} (Hint: use prime factorization and the previous problem along with results earlier in the chapter.)

4.

Clearly 4 has a square root modulo 7\text{.} Find all square roots of 4 modulo 7^3 without using Sage or trying all 343 possibilities. Why is this exercise not as challenging as it seems, and what would you do to make it harder?

5.

Solve x^2+3x+5\equiv 0\text{ (mod }15) using completion of squares and trial and error for square roots.

Solve the following congruences without using a computer.

6.

x^2+6x+5\equiv 0 (mod 17)

7.

5x^2+3x+1\equiv 0 (mod 17)

8.

Prove that if p is an odd prime

\begin{equation*} \sum_{a=1}^{p-1}\left(\frac{a}{p}\right)=0\text{.} \end{equation*}
9.

Explore and conjecture a formula for

\begin{equation*} \sum_{a\in Q_p} a\text{,} \end{equation*}

possibly dependent upon some congruence class for p\text{.}

10.

Show that a quadratic residue can't be a primitive root if p>2\text{.}

11.

Show that if p is an odd prime, then there are exactly \frac{p-1}{2}-\phi(p-1) residues which are neither QRs nor primitive roots. (Hint: don't think too hard – just do the obvious counting up.)

13.

Evaluate Legendre symbols for all a\neq 0 where p=7\text{,} using Euler's Criterion.

14.

Explore for a pattern for when -5 is a quadratic residue. Try not to use any fancy criteria, but just to seek a pattern based on the number.

15.

Use Euler's Criterion and the ideas of Proof 16.7.1 to prove that 3 has a square root modulo p if p\equiv 1\text{ (mod }12)\text{.} (See Proposition 17.3.4 for full details of \left(\frac{3}{p}\right)\text{.})

16.

Explore for a pattern for, given p\text{,} how many pairs of consecutive residues are both actually quadratic residues. Then connect this idea to the following formula, which you should evaluate for the same values of p\text{:}

\begin{equation*} \sum_{a=1}^{p-2}\left(\frac{a}{p}\right)\left(\frac{a+1}{p}\right) \end{equation*}

(A harder problem is to prove your evaluation works for all p\text{.})