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Solutions to Assignment 2
September 16, 2009
Section 2.1 2. A complete residue system modulo 17 can be taken to be
{− 8 , − 7 , · · · , − 1 , 0 , 1 , · · · , 7 , 8 }
For each element, we can add suitable multiples of 17 to make it divisible by 3. One possible answer is:
{ 9 , − 24 , − 6 , 12 , − 21 , − 3 , 15 , − 18 , 0 , 18 , − 15 , 3 , 21 , − 12 , 6 , 24 , − 9 }
- We check the values of n^2 mod 10 for the complete residue system modulo 10 given by {− 4 , − 3 , · · · , 0 , 1 , · · · , 4 , 5 }. Clearly it suffices to check the following:
02 ≡ 0 mod 10 12 ≡ 1 mod 10 22 ≡ 4 mod 10 32 ≡ 9 mod 10 42 ≡ 6 mod 10 52 ≡ 5 mod 10
as stated.
- Note that 42 = 2 × 3 × 7. Now by Fermat’s Little Theorem,
n^7 ≡ n mod 7 (1)
for any integer n. On the other hand, the congruences
n^7 ≡ n mod 3 (2)
and
n^7 ≡ n mod 2 (3)
is easily checked to be valid for any n (for example, to check (2), it suffices to evaluate both sides on a c.r.s. modulo 3, say given by {− 1 , 0 , 1 }). Combining these three congruences, we obtain n^7 ≡ n mod 42.
- For n ≥ 1. Argue by contradiction. Suppose that p 1 , · · · , pn are all the primes of the form 4k + 1. Form the number:
N = (2p 1 · · · pn)^2 + 1 > 1
Let q be a prime dividing N (which must be odd). Then we would have
(2p 1 · · · pn)^2 + 1 ≡ 0 mod q
i.e. (2p 1 · · · pn)^2 ≡ −1 mod q
By theorem 2.12, we must have q ≡ 1 mod 4. This implies that q must be one of the p 1 , · · · , pn. But then we obtain a contradiction as in the proof of the infinitude of primes. Section 2. 5(e). One can directly apply the Euclidean Algorithm to find an inverse of 64 mod 105 to solve the congruence. However, we can also proceed as follows: notice that 105 = 3 × 5 × 7. By Chinese Remainder Theorem, it’s equivalent to solve
64 x ≡ 83 mod 3 64 x ≡ 83 mod 5 64 x ≡ 83 mod 7
which simplifies to:
x ≡ 2 mod 3 x ≡ −3 mod 5 (4) x ≡ −1 mod 7 (5)
Using the algorithm of the Chinese Remainder Theorem: we solve:
35 c 1 ≡ 1 mod 3 21 c 2 ≡ 1 mod 5 15 c 3 ≡ 1 mod 7
One may take:
c 1 = − 1 c 2 = 1 c 3 = 1
From this, one obtains the solution to the system (4) by the formula
x ≡ 2 × 35 × −1 + − 3 × 21 × 1 + − 1 × 15 × 1 mod 105 ≡ − 148 ≡ 62 mod 105
