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98 3. Factors and Zeros

(f) A second solution of the congruence t2 + 2 G 0 (mod 3) is t z 2

(mod 3). Find a solution of t2 + 2 - 0 (mod 243) of the form

t = 3k -t 2.

Determine the number of incongruent solutions for each of the fol-

lowing congruences:

(a) 2t3 + t + 3 E 0 (mod 1000)

(b) 2t3 + t + 3 s 0 (mod 83349).

Solve the congruence: 3x3 - 4x2 + 103: - 3 E 0 (mod 675). For one

of the solutions, evaluate the polynomial and verify that the value is

divisible by 675.

Consider the polynomial equation

t5 - 4t4 - 411t3 - 452t2 - 3322t - 828 = 0.

12

13.

14.

15.

Explorations

(a) Argue that the absolute value of any root is less than 5000.

(b) Find those values of t between -5000 and 5000 for which the

value of the polynomial is a multiple of 5000. Argue that any

integer root of the equation must be one of these values and thus

find all its integer roots.

Using congruences, find all integer roots of the equation

2t4 + 20t3 + 19t2 - t - 90 = 0.

E.32. Little Fermat Theorem. Verify that n3 E n (mod 3) and n5 E n
(mod 5) for each integer n. These are particular instances of a general
result:
np s n (mod p)

whenever p is prime and n is an integer.
One way to see this is to follow an argument given by L. Euler (1707-
1783). Since the result is clear for p = 2, we suppose that p is an odd prime

and prove the result by induction on n. Observe that P
( >

k is a multiple

ofpforl<k<p-1.
The case n = 1 is obvious. Euler gave two arguments for the n = 2 case.
First, expand the right side of 2P = (1 + 1)P binomially and write as a
congruence modulo p. Alternatively, one can get a slightly stronger result
by using

2P-’ - 1= (1+ 1)P-’ - 1= pg ( p r, l )

k=l