Download 3. Prove that n3 + (n + 1)

3. Prove that n3 + (n + 1)3 + (n + 2)3 ≡ 0(mod9) for all n - AK
Solution 1: Plug in n = 0, 1, ..., 7, 8 and verify it. This works but defeats the
purpose of what we’re trying to learn
Solution 2: Simplify equation to: 3n3 + 9n2 + 15n + 9, and simplify mod 9 to
3
3n + 6n = 3n(n2 + 2). We’ve reduced the problem to showing that 3n(n2 + 2) ≡
0(mod9), or n(n2 + 2) ≡ 0(mod3). Again, this can be verified by checking for
n = 0, 1, 2. This is better than Solution 1, but even better:
Solution 2a: Notice that n2 + 2 cannot be factored in the integers, but (!)
2 ≡ −1(mod3), so n2 + 2 ≡ n2 − 1 = (n + 1)(n − 1). Now we see that
n(n2 + 2) ≡ n(n + 1)(n − 1)(mod3), and this is clearly zero since 3 must divide
the product of three consecutive integers.
4 - Solve the following set of simultaneous congruences...
The relevant congruences will be solved whenever x ≡ 251 (mod 360).
5 - How many square numbers are there mod p? - AK
A square number in general is a number, n, such that n is the square of some
other number. I.e. n is a square number if there exists a number x such that
x2 = n. We’re all familiar with the usual squares, 0, 1, 4, 9, 16,... (Why did many
of you insist on leaving out zero? It’s as good a number as any other, if not better!)
In the language of modular arithmetic, this simply means that n is a square mod m
if there exists an x such that x2 ≡ n(modm). E.g. In mod 7, 32 = 9 ≡ 2(mod7),
so 2 is a square number (of course, it’s not in the integers themselves). Compute
the squares for the first few primes to guess the pattern that the first half repeats
in reverse order in the second half. Then there are two things to prove. Namely,
we have to show that k and p − k have the same square mod p, and that no other
number has the same square. We kill both birds with one stone by supposing that
a2 ≡ b2 (modp). Then a2 − b2 ≡ 0(modp), or (a + b)(a − b) ≡ 0(modp). Since
p is prime, if p divides a product of two terms, it must divide at least one of them,
so we have a + b ≡ 0 or a − b ≡ 0. This corresponds to the cases a ≡ −b ≡ p − b
have distinct
or a ≡ b, respectively. So zero is a square, and then 1 through p−1
2
p+1
squares, and these repeat in reverse order from 2 through p − 1. Thus there are
p−1
+ 1 = p+1
squares mod p.
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6 and 7 - These two results combine to form what is called Wilson’s Theorem.
This says that if n is a prime number then (n − 1)! ≡ −1 (mod n) and if n is
composite then it is not (in particular it will be congruent to 0 as long as n > 4.)
Assume p is prime. Then we know that for all a < p there exists a unique
value of a0 such that aa0 ≡ 1 (mod p). However, we also proved in class that
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a2 ≡ 1 (mod p) if and only if a ≡ 1 or a ≡ −1. In particular, this shows that for
all a other than 1 and p − 1 we get that a0 is different from a. This allows us to
break all the numbers from 2 to p − 2 into pairs whose product are 1. However,
this in turn implies that
(p − 1)! ≡
≡
≡
≡
(1)(2) . . . (p − 2)(p − 1)
(1)(a1 )(a01 ) . . . (ar )(a0r )(p − 1)
(1)(1) . . . (1)(p − 1)
−1
It may help to think through an example. For example, if p = 7 then I am
claiming that the numbers between 2 and 5 break up into pairs whose product is
1. But this is clear as 2 ∗ 4 ≡ 3 ∗ 5 ≡ 1 (mod 7).
If n is a composite number which is not of the form p2 for some prime then n
can be factored as n = rs where r and s are distinct numbers less than n. In this
case, n = rs|(n − 1)! and so in particular (n − 1)! ≡ 0 (mod n).
Finally, if n is of the form p2 then we note that as long as p > 2, then p and
2p are distinct numbers less than (n − 1), so we have that n|p(2p)|(n − 1)! and
therefore (n − 1)! ≡ 0 (mod n).
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