Mathematical Reasoning- Writing and Proof, Version 2.1, 2014a

3.5. The Division Algorithm and Congruence 151
0; 1; 2; : : :; or n 1 modulo n. For example, if we assume that 5 does not divide an
integer a, then we know a is not congruent to 0 modulo 5, and hence, that a must
be congruent to 1, 2, 3, or 4 modulo 5. We can use these as 4 cases within a proof.
For example, suppose we wish to determine the values of a 2 modulo 5 for integers
that are not congruent to 0 modulo 5. We begin by squaring some integers that are
not congruent to 0 modulo 5. We see that
1 2 D 1 and 1 1.mod 5/ :
3 2 D 9 and 9 4.mod 5/ :
6 2 D 36 and 36 1.mod 5/ :
8 2 D 64 and 64 4.mod 5/ :
9 2 D 81 and 81 1.mod 5/ :
These explorations indicate that the following proposition is true and we will now
outline a method to prove it.
Proposition 3.33. For each integer a, ifa 6 0.mod 5/, thena 2 1.mod 5/ or
a 2 4.mod 5/.
Proof. We will prove this proposition using cases for a based on congruence modulo
5. In doing so, we will use the results in Theorem 3.28 and Theorem 3.30.
Because the hypothesis is a 6 0 .mod 5/, we can use four cases, which are: (1)
a 1.mod 5/, (2)a 2.mod 5/, (3)a 3.mod 5/, and(4)a 4.mod 5/.
Following are proofs for the first and fourth cases.
Case 1. .a 1.mod 5//. In this case, we use Theorem 3.28 to conclude that
a 2 1 2 .mod 5/ or a 2 1.mod 5/ :
This proves that if a 1.mod 5/,thena 2 1.mod 5/.
Case 4. .a 4.mod 5//. In this case, we use Theorem 3.28 to conclude that
a 2 4 2 .mod 5/ or a 2 16 .mod 5/ :
We also know that 16 1 .mod 5/. So we have a 2 16 .mod 5/ and 16
1 .mod 5/, and we can now use the transitive property of congruence (Theorem
3.30) to conclude that a 2 1 .mod 5/. This proves that if a 4 .mod 5/,
then a 2 1.mod 5/.