Consider congruence mod 5. Let r = {0, 1, 2, 3, 4}. 1) Find r so that 3 =r mod (5). 2) Find r so that 32 = r mod (5). 3) Find r so that 33 = r mod (5) 4) Find r so that 34 = r mod (5). 5) Find r so that 35 = r mod (5).

1. For each solution. JX² = -1 (mod 397) (397 is prime) CO congruence determine whether there is a x² +8x + 14 = 0 (mod 37) ししし iii) x² = 785 (mod 787) (787 is prime)

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6. Give the addition and multiplication tables for nonconstant polynomials in Z3[x]/(x² + x+2). (i.e. just the linear polynomials x, x+1, x+2, 2x, 2x + 1, 2x + 2) Hint: By the mod x² = -x-2 (mod x² + x + 2) = 2x + 1 (mod x² + x + 2).

Let p be an odd prime and let a and b be integers with ab #0(mod p). Show that ax² = b[mod p) has a solution if and only if ()) = 1.

How can I prove this?