Download Elementary Number theory and Cryptography Epiphany 2011

Elementary Number theory and Cryptography
Epiphany 2011, Solutions to Problem Sheet 3
1. Represent each of the primes 113, 229, 373 as a sum of two
squares.
113 = 72 + 82 ; 229 = 152 + 22 ; 373 = 72 + 182 .
2. Prove that 2n , where n ∈ N, is a sum of two squares.
If n = 2m is even then 2n = (2m )2 + 02 .
If n = 2m + 1 is odd then 2n = (2m )2 + (2m )2 .
3. Prove that if n is congruent to 3 or 6 modulo 9 then n can’t be
presented as a sum of two squares.
Suppose n = a2 + b2 with a, b ∈ Z and n ≡ 3 or 6 mod 9.
Prove that both a and b are not divisible by 3. Indeed, suppose
a = 3a1 with a1 ∈ Z. Then b2 = n − a2 ≡ 0 mod 3 implies that b = 3b1
with b1 ∈ Z. Therefore, a2 + b2 = 9(a21 + b21 ) ≡ 0 mod 9. Contradiction.
Therefore, a and b are congruent to 1 or 2 modulo 3. This implies
that both a2 and b2 are congruent to 1 modulo 3 and a2 +b2 ≡ 2 mod 3.
Contradiction.
4. A natural number is called “triangular” if it equals a(a + 1)/2 for
some a ∈ N. Prove that if n is a sum of two triangular numbers then
4n + 1 is a sum of two squares.
Suppose n = a(a + 1)/2 + b(b + 1)/2 with a, b ∈ Z. Then 4n + 1 =
2a(a + 1) + 2b(b + 1) + 1 = (a2 − 2ab + b2 ) + (a2 + b2 + 2ab + 2a + 2b + 1) =
(a − b)2 + (a + b + 1)2 .
5. Prove that a prime number p can be written as a sum of two
squares if and only if the congruence x2 + 1 ≡ 0 mod p admits a solution.
If p = a2 + b2 with a, b ∈ N, then a2 + b2 ≡ 0 mod p with b 6≡
0 mod p. Dividing this congruence by b mod p we obtain that (a/b)2 ≡
−1 mod p. This means that −1 is a quadratic residue modulo p. Inversely, if -1 is a quadratic residue then p ≡ 1 mod 4 and p is a sum of
two squares.
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6. Find a natural number n having at least three different representations as the sum of two squares of natural numbers, disregarding signs
and the order of summands.
Take p = 5·11·17 = 1105. Then 5, 13 and 17 are sums of two squares
(they all are primes ≡ 1 mod 4). More specifically, for the Gaussian
integers α1 = 2 + i, α2 = 3 + 2i and α3 = 4 + i, we have α1 ᾱ1 = 5,
α2 ᾱ2 = 13 and α3 ᾱ3 = 17. Then we can find different presentations
1105 as sums of two squares by the use of different “grouppings” in the
product α1 ᾱ1 α2 ᾱ2 α3 ᾱ3 = 1105 as follows.
— (α1 α2 α3 )(α1 α2 α3 ) = 1105, then α1 α2 α3 = 9 + 32i and 92 + 322 =
1105.
— Here the product of α1 α2 ᾱ3 and its complex conjugate again gives
1105; then α1 α2 ᾱ3 = 23 + 24i and 232 + 242 = 1105.
— Here the product of α1 ᾱ2 ᾱ3 and its complex conjugate gives 1105;
then α1 ᾱ2 ᾱ3 = 33 + 4i and 332 + 42 = 1105.
7. Prove that the natural number n has as many representations as
the sum of two integer squares as does the integer 2n.
Note that n = a2 +b2 with a, b ∈ Z if and only if for α = a+bi ∈ Z[i],
it holds αᾱ = n. In particular, 2 = 12 +12 means that (1+i)(1 + i) = 2.
Then for β = α(1+i) , it holds β = (a+bi)(1+i) = (a−b)+(a+b)i, one
has 2n = β β̄ = (a−b)2 +(a+b)2 . This allows to attach to any sum of two
squares n = a2 + b2 for n the sum of two squares 2n = (a − b)2 + (a + b)2
for 2n. It remains to verify that any presentation 2n = c2 + d2 , where
c, d ∈ Z, can be obtained in this way. Clearly, c ≡ d mod 2 (i.e. c
and d are either both odd or even). Therefore, we can introduce the
integers a = (c − d)/2 and b = c + d/2 and our presentation for 2n
comes from the presentation n = a2 + b2 .
8. Prove that of any four consecutive natural numbers, at least one
is not representable as a sum of 2 squares.
Suppose n = a2 + b2 is the sum of two square with a, b ∈ Z. Note
that a2 or b2 can be congruent only to 0 or 1 modulo 4. Therefore, n
can be congruent only to 0,1 or 2 and can’t be congruent to 3 modulo
4. But one of four consecutive integers will be always congruent to 3
modulo 4 and, therefore, can’t be written as a sum of two squares.
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9. If a prime number is the sum of two or four squares of prime
numbers then one of these primes must be equal to 2.
Suppose p = p21 + p22 , where all p, p1 , p2 are prime numbers. If neither
of p1 and p2 equals 2 then they are both odd, p = p21 + p22 is even and
> 18. Contradiction.
Similarly, if p = p21 + p22 + p23 + p24 , where all p, p1 , .p2 , p3 , p4 are primes
and neither of p1 , p2 , p3 , p4 equals 2 then they all are odd and p is even
and > 36. Contradiction.
10. If a prime number is the sum of three squares of prime numbers
then one of these primes must be equal to 3.
Suppose p = p21 + p22 + p23 , where p, p1 , p2 , p3 are primes. If neither of
p1 , p2 , p3 equals three then p21 ≡ p22 ≡ p23 ≡ 1 mod 3 and p is divisible
by 3, but p > 12.
11. Let p be an add prime. If p divides a2 + b2 where a and b are
relatively prime natural numbers then prove that p ≡ 1 mod 4. Deduce
from this that any non-trivial divisor of a sum of two relatively prime
squares is again of two squares.
Note that b 6≡ 0 mod p (otherwise a ≡ 0 mod p and gcd(a, b) 6= 1).
Then a2 + b2 ≡ 0 mod p implies that (a/b)2 ≡ −1 mod p, −1 is a
quadratic residue modulo p and p ≡ 1 mod 4. Therefore, all prime
divisors of a2 + b2 are congruent to 1 modulo 4. This implies that any
divisor of a2 + b2 satisfies the same condition and is therefore a sum of
two squares.
12. Prove that every prime number p of the form 8k + 1 or 8k + 3
can be written in the form p = a2 + 2b2 .
If p = 8k + 1 then both −1 and 2 are quadratic residues modulo p.
If p = 8k + 3 then both −1 and 2 are not quadratic residues modulo
p. Therefore, in both above cases −2 is a quadratic residue modulo p.
Now mimic the proof of Theorem that p = a2 + b2 if p ≡ 1 mod 4 (or
equivalently, if −1 is a quadratic residue modulo p).
Let α ∈ Z be such that α2 is congruent to -2 modulo p. Then
x2 + 2y 2 ≡ (x − αy)(x + αy) mod p.
√
Consider the set S = {(x0 , y0 ) | 0 6 x0 , y0 < p}.
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√ √
Then S contains more than p · p = p elements and there are two
different elements (x1 , y1 ), (x2 , y2 ) ∈ S such that
x1 − αy1 ≡ x2 − αy2 mod p
√
Therefore, for x0 = x1 −x2 and y0 = y1 −y2 , we have − p < x0 , y0 <
√
p, (x0 , y0 ) 6= (0, 0), and x20 + 2y02 ≡ (x0 − αy0 )(x0 + αy0 ) ≡ 0 mod p.
But this means that 0 < x20 + 2y02 < 3p and, therefore, x20 + 2y02
equals either p or 2p. In the first case our problem is solved. In the
second case x0 is divisible by 2 and substituting x0 = 2x1 we obtain
y02 + 2x21 = p.
13. A natural number m is representable as difference of squares of
two integers if and only if it is the product of two factors that are both
even or both odd.
Suppose m = a2 − b2 with a, b ∈ Z. Then m = (a − b)(a + b),
where a − b ≡ a + b mod 2. Inversely, if m = m1 m2 , m1 , m2 ∈ Z
and m1 ≡ m2 mod 2 then a = (m1 + m2 )/2 and b = (m1 − m2 )/2 are
integers and m = a2 − b2 .
14. An even natural number can be written as the difference of two
squares if and only if it is divisible by 4.
This follows immediately from above problem 13.
15. Find three different representations of 45 as the difference of
two squares. Explain why 45 is the smallest natural number with this
property.
Use three different ordered factorisations 45 = 1 · 45 = 3 · 15 = 5 · 9
to obtain that 45 = 232 − 222 = 92 − 62 = 72 − 22 .
16. Establish that the equation a2 + b2 + c2 + a + b + c = 1 has no
integer solutions.
Multiply both sides by 4 and complete squares to get (2a + 1)2 +
(2b + 1)2 + (2c + 1)2 = 7. Use that the left hand side is congruent to 3
modulo 8.
17. Prove that no integer of the form 9k + 4 or 9k + 5 can be a sum
of three or fewer cubes.
Note that if an integer a ≡ 0 mod 3 then a3 ≡ 0 mod 9 and if
a ≡ ±1 mod 3 then a3 ≡ ±1 mod 9. Therefore, if N = a3 + b3 + c3 for
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a, b, c ∈ Z then N can only congruent to 0, ±1, ±2 modulo 9. Therefore,
it can’t be congruent to 4 or 5 modulo 9.
18. Prove that the only prime which can be represented as a sum of
two positive cubes is 2.
If p = a3 + b3 with a, b ∈ N then p = (a + b)(a2 − ab + b2 ) implies
that a + b = p and a2 − ab + b2 = 1. But the second condition gives
(a − b)2 + ab = 1, ab 6 1 and therefore a = b = 1.
19. Express each of the primes 7, 19, 37, 61 and 127 as the difference
of two cubes.
If p = a3 −b3 then a and b could be found from the relations a−b = 1
and a2 + ab + b2 = p. Then:
— if p = 7 then a = 2 and b = 1;
— if p = 19 then a = 3 and b = 2;
— if p = 37 then a = 4 and b = 3;
— if p = 61 then a = 5 and b = 4;
— if p = 127 then a = 7 and b = 6.
20. Verify the formula
(a21 + a22 + a23 + a24 )(b21 + b22 + b23 + b24 ) =
(a1 b1 − a2 b2 − a3 b3 − a4 b4 )2 + (a2 b1 + a3 b4 − a4 b3 + a1 b2 )2 +
(a3 b1 + a1 b3 − a2 b4 + a4 b2 )2 + (a4 b1 + a1 b4 + a2 b3 − a3 b2 )2
Use this formula to deduce that if m, n ∈ N are sums of four squares
then mn also has this property.
The required formula follows from the formula for multiplication of
two quaternions: (a1 + a2 i + a3 j + a4 k)(b1 + b2 i + b3 j + b4 k) =
(a1 b1 − a2 b2 − a3 b3 − a4 b4 ) + (a2 b1 + a3 b4 − a4 b3 + a1 b2 )i+
(a3 b1 + a1 b3 − a2 b4 + a4 b2 )j + (a4 b1 + a1 b4 + a2 b3 − a3 b2 )k
21. Express the integers
a) 231 = 3 · 7 · 11;
b) 391 = 17 · 23;
c) 2109 = 37 · 57
as sums of four squares.
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a) Take the quaternions α = 1 + i + j, β = 2 + i + j + k and
γ = 3 + i + j. They correspond to the presentations of 3, 7 and 11
as sums of 4 squares, in other words 3 = αᾱ, 7 = β β̄, 11 = γγ̄.
Then compute αβγ = −6 + 11i + 7j + 5k. This gives the presentation
62 + 112 + 72 + 52 = 231 = 3 · 7 · 11.
b) Similarly, use α = 4+i, β = 3+3i+2j+k and αβ = 9+15i+7j+6k
to get the presentation 92 + 152 + 72 + 62 = 391.
c) Use (6 + i)(7 + 2i + 2j) = 40 + 19i + 12j + 2k to get that
2109 = 402 + 192 + 122 + 22 .