SOLUTIONS FOR REVISION LECTURE QUESTIONS
Solution (A1).
(a) The two square theorem says: a positive integer n is a
sum of 2 squares if and only if every prime p ≡ 3 (mod 4) which divides n
appears with an even power in the prime factorisation of n.
(b) Consider the prime factorisations of m and n. We have m = pa1 1 pa2 2 · · · par r
br+1
· · · pbss , where, since m|n, we have bi ≥ ai for
and n = pb11 pb22 · · · pbrr pr+1
i = 1, . . . , r.
Since m, n are both sums of two squares, the two square theorem says
that ai and bi are even if pi ≡ 3 (mod 4). So bi − ai is even too. We have
b
r+1
· · · pbss
n/m = pb11 −a1 pb22 −a2 · · · pbrr −ar pr+1
so by the two square theorem n/m is a sum of two squares.
Solution (A2).
(a) If we consider the equation mod 2 there are no solutions:
x ≡ 0 or x ≡ 1 (mod 2) implies that x2 + 3x + 7 ≡ 1 (mod 2). So there are
no solutions mod 6.
(b) It suffices to show that there are no solutions modulo 7. By Fermat’s little
theorem, z 6 + 3 ≡ 3 or 4 (mod 7). By Euler’s criterion, (or because they
square to 0 or 1) x3 , y 3 are ≡ 0, 1 or −1 (mod 7). So x3 + y 3 ≡ 0, 1, −1, 2
or −2 (mod 7). So there are no common values for the two sides of the
equations modulo 7, hence no solutions.
Solution (A4).
(a) This means that α is a root of a non-zero polynomial with
rational coefficients.
(b) This means that α is not an algebraic number.
(c) Suppose log10 5 = a/b, with a ∈ Z and b ∈ N. Then 10a/b = 5 so 10a = 5b ,
with b ≥ 1. But this contradicts the fundamental theorem of arithmetic.
Solution (B8).
(a) pm implies that (n!)4 ≡ −1 (mod p). Which implies that
2 2
((n!) ) ≡ −1 (mod p), so we can take a = (n!)2 .
(b) a2 ≡ −1 (mod p) implies that a2 ≡ (n!)4 (mod p) so a ≡ ±(n!)2 (mod p).
(n!)2 2
So ap = ±(n!)
= ±1
= ±1
by multiplicativity of the Legendre
p
p
p
p
2
symbol. Since a ≡ −1 we know that −1
= +1 so in either case we get
p
a
=
+1.
p
p−1
Now Euler’s criterion implies that a 2 ≡ 1 (mod p). Since −1
= +1,
p
p−1
p−1
2
we know that p ≡ 1 (mod 4). So p−1
= (a2 ) 4 ≡
4 is an integer and a
p−1
p−1
(−1) 4 (mod p). Putting everything together, we get (−1) 4 = 1.
(c) Suppose for a contradiction that every prime of the form 8k + 1 is ≤ n,
for some positive integer n. Consider m = (n!)4 + 1, and let p|m be a
prime divisor of m. Since (n!)4 ≡ −1 (mod p), p > n. This means that
p 6≡ 1 (mod 8), by our assumption. On the other hand, part (b) implies that
p−1
(−1) 4 = 1 which implies that p ≡ 1 (mod 8). So we have a contradiction,
which proves that there are infinitely many primes of the form 8k + 1.
Solution (B3).
(a) If n is a positive integer φ(n) is the number of elements
of the set Z×
n . Equivalently, it is the number of integers between 1 and n
which are coprime to n.
Date: Thursday 27th April, 2017.
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SOLUTIONS FOR REVISION LECTURE QUESTIONS
(b) Since m, n are coprime the Chinese remainder theorem says that there is a
bijection between Zmn and Zm × Zn , given by taking [a]mn to ([a]m , [a]n ).
Moreover, a is coprime to mn if and only if it is coprime to m and n (e.g. by
considering the prime divisors of m and n, or by a result from lectures),
×
×
so we get a bijection between Z×
mn and Zm × Zn . Comparing the sizes of
these two sets gives the desired equality.
(c) Applying part (b), it suffices to show that for each i we have φ(pai i ) =
pai i − pai i −1 . To show this, we observe that an integer is coprime to pai i if
and only if it is coprime to pi . So the integers between 1 and pai i which are
coprime to pai i are everything except the pai i −1 multiples of p. Therefore
the total number is pai i − pai i −1 .
(d) Since p is the smallest prime that does not divide n, all the prime divisors
of p − 1 do divide n. If n = pa1 1 · · · par r with ai > 0 then p − 1 = pb11 · · · pbrr
with bi ≥ 0. Applying part (c) twice we get
!
Y
Y
Y
ai +bi −1
bi
ai +bi
− pi
)=
pi
(pai i − pai i −1 ) = (p − 1)φ(n).
φ((p − 1)n) =
(pi
i
i
i
Solution (B4).
(a) The necessary and sufficient condition is that d = gcd(a, m)
divides b.
(b) The solutions are
x ≡ [a/d]−1
m/d [b/d]m/d
(mod m/d)
So there is 1 solution (mod m/d) and hence d solutions (mod m).
(c) We have d = gcd(12, 16) = 4. 4|8 so there are 4 solutions mod 16.
3
(d) Since 17 − 1 = 24 , it suffices to show that 32 6≡ 1 (mod 17), as the order
of 3 is a divisor of 24 . We have 38 = (81)2 ≡ (−4)2 ≡ −1 (mod 17). So 3
is a primitive root mod 17.
(e) If 17|b then we have 1 solution, x ≡ 0 (mod 17). If 17 - b then we have
b ≡ 3j (mod 17) for some j and we can write x ≡ 3i (mod 17). Then
the equation becomes 312i ≡ 3j (mod 17), which is equivalent to 12i ≡ j
(mod 16). By parts (a) and (b) this has either 4 or 0 solutions, depending
on whether 4|j or not.
Solution (B6).
(a) The order of g modulo p is equal to p − 1.
(b) ap = +1 if a is a quadratic residue mod p and −1 is a is not a quadratic
residue mod p. If p|a we also define ap = 0.
(c) If p, q are distinct odd primes then pq = − pq if p ≡ q ≡ 3 (mod 4), and
p
q
q = p otherwise.
(d) Suppose m = 2n m0 with m0 odd. Then
n
n
2m + 1 = (22 + 1)(22
(m0 −1)
n
− 22
(m0 −2)
n
+ · · · − 22 + 1)
so m0 > 1 implies that 2m + 1 is not prime. Hence if 2m + 1 is prime we
must have m0 = 1 and m is a power of 2.
n
(e) 3 is a primitive root if and only its order is 22 . This happens if and only
2n
if 32 −1 = 3(p−1)/26≡ 1 (mod p). By Euler’s criterion, this happens if and
only if p > 3 and p3 = −1. Now we apply the quadratic reciprocity law. If
2n
p > 3 then n ≥ 1, so p ≡ 1 (mod 4). So p3 = p3 = (−1)3 +1 = 23 = −1.
n
So 3 is a primitive root mod p for all n > 0 with p = 22 + 1 prime.