Modular Arithmetic Recap We use the notation an...a2a1a0 for the ...

Modular Arithmetic Recap
We use the notation a n ...a 2 a 1 a 0 for the number formed with the digits a 0 , a 1 , ..., a n ,
so that:
a n ...a 2 a 1 a 0 = a 0 + 10a 1 + 100a 2 + ... + 10 n a n .
Fix a number p. By reducing each power of 10 modulo p, we can find simplified ways
of writing a n ...a 2 a 1 a 0 ( mod p) .
Thus:
a n ...a 2 a 1 a 0 ≡ a 0 ( mod 2)
because 10 k ≡ 0 ( mod 2) for all k ≥ 1.
a n ...a 2 a 1 a 0 ≡ a 0 + a 1 + a 2 + ... + a n ( mod 3)
because 10 k = 99...9 + 1 ≡ 1 ( mod 3) for all k ≥ 1.
a n ...a 2 a 1 a 0 ≡ a 1 a 0 ( mod 4)
because 10 k ≡ 0 ( mod 4) for all k ≥ 2.
a n ...a 2 a 1 a 0 ≡ a 0 ( mod 5)
because 10 k ≡ 0 ( mod 5) for all k ≥ 1.
a n ...a 2 a 1 a 0 ≡ 4(a 0 + a 1 + a 2 + ... + a n ) ( mod 6)
because 10 k ≡ 4
( mod 6) for all k ≥ 1 (Prove this by induction).
a n ...a 2 a 1 a 0 ≡ a 0 + 3a 1 + 2a 2 − a 3 − 3a 4 − 2a 5 + a 6 + 3a 7 + 2a 8 ...) ( mod 7)
because
10 ≡ 3 ( mod 7) =⇒
10 2 ≡ 3 2 ≡ 2 ( mod 7) =⇒
10 3 ≡ 2 × 3 ≡ −1 ( mod 7) =⇒
10 4 ≡ −1 × 3 ≡ −3 ( mod 7) =⇒ ...
a n ...a 2 a 1 a 0 ≡ a 2 a 1 a 0 ( mod 8)
because 10 k ≡ 0 ( mod 8) for all k ≥ 3.
a n ...a 2 a 1 a 0 ≡ a 0 + a 1 + a 2 + ... + a n ( mod 9)
because 10 k = 99...9 + 1 ≡ 1 ( mod 9) for all k ≥ 1.
a n ...a 2 a 1 a 0 ≡ a 0 ( mod 10).
a n ...a 2 a 1 a 0 ≡ a 0 − a 1 + a 2 + ... + (−1) n a n ( mod 11)
because 10 k =≡ (−1) k ( mod 11) for all k ≥ 0.
a n ...a 2 a 1 a 0 ≡ a 2 a 1 a 0 − a 5 a 4 a 3 + a 8 a 7 a 6 − ... ( mod 1001)
because 10 3k ≡ (−1) k ( mod 1001) for all k ≥ 3.
This last equality can prove useful when you work mod 7, 11 or 13 because
1001 = 7 × 11 × 13 .
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Perfect squares
Which numbers n can be perfect squares?
Note: There are only p equivalence classes (mod p): n = 0, 1, 2, ..., or p − 1.
As n 2 ≡ (−n) 2 ≡ (p − n) 2 ( mod p), , there are at most p/2 possible values that
a perfect square can take mod p:
n 2 ≡ 0 or 1 ( mod 3),
because 02 ≡ 0 ( mod 3),
1 2 ≡ 2 2 ≡ 1 ( mod 3).
n 2 ≡ 0 or 1 ( mod 4),
because 02 ≡ 2 2 ≡ 0 ( mod 4),
1 2 ≡ 3 2 ≡ 1 ( mod 4).
n 2 ≡ 0 or 1 or −1 ( mod 5),
because 12 ≡ 4 2 ≡ 1 ( mod 4),
2 2 ≡ 3 2 ≡ −1 ( mod 4),
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n 2 ≡ 0 or 1 or 4 or 5 or 6 or 9 ( mod 10).
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If p is prime, then p 2 ≡ 1 ( mod 12) .
Indeed, looking at the factors of 12k, 12k +2, 12k +3, 12k +4, 12k +6, 12k +8, 12k +
9, 12k + 10 we see these can’t be written as p 2 for p prime.
Then 12k + 5 ≡ 12k + 11 ≡ 2 ( mod 3) so they can’t be perfect squares, and
12k + 7 ≡ 3 ( mod 4) so it can’t be a perfect square.