Download Homework 3 Solutions

Homework 3 Solutions
1. We proceed with the following algebraic manipulation:
x2 − 3x − 1 ≡ 0
(mod 31957)
x2 − 3x ≡ 1
(mod 31957)
x − 3x + (3 · 2) ≡ 1 + (3 · 2)2
2
2
2
(x − 3 · 2) ≡ 1 + 9 · 4
(mod 31957)
(mod 31957)
≡ 4(4 + 9) ≡ 4 · 13
(mod 31957)
So it is clear from this that such an x-value can be found if 4 · 13 is a
square mod 31957—or, equivalently, if 13 is a square mod 31957 (since
2
clearly 4 = 2 ). This reduces simply to taking the Legendre symbol
13
31957
3
16
=
=
=
=1
31957
13
13
13
where we have used Quadratic Reciprocity along with the fact that
31957 is 1 mod 4.
(Remark 1: The “only if” version—that there would be no solution if
the Legendre symbol were −1—is true, but we don’t need it here.)
(Remark 2: If we were looking for roots in the real numbers, instead of
modulo a prime, we’d use the quadratic formula and find that the discriminant of the quadratic is 13. That is essentially the same method
as this one, where, if you’ll notice, all we’re doing is completing the
square. The Legendre symbol calculation is only meant to check that
“the square root of 13” makes sense modulo 31957.)
2. First, let’s show there are infinitely many primes congruent to 2 mod
3. Suppose for a contradiction that there were only finitely many.
Label them p0 , p1 , . . . , pn , with p0 = 2. Now consider the number
N = 3 · p1 · p2 · · · pn + 2. Notice that N is odd, because it is the product of several odd numbers plus an even number. That means it is not
divisible by 2. Also notice that N ≡ 2 (mod 3), so it is not divisible
by 3. Finally, notice that no odd prime congruent to 2 mod 3 divides
N , since all those primes are included in the product 3 · p1 · p2 · · · pn ,
so if one of those primes were to divide N , it would also divide 2,
1
which is impossible. Now if 3 doesn’t divide N , and no prime congruent to 2 mod 3 divides N , then all the prime divisors of N must
be 1 mod 3. But this is a contradiction, because then any product of
these primes—in particular, N —is 1 mod 3, yet N is 2 mod 3. So we
have shown that there are infinitely many primes congruent to 2 mod 3.
For the next part, suppose for a contradiction that there are only
finitely many primes congruent to 1 mod 3. As directed, label them
p1 , . . . , pk and consider N = (2p1 p2 · · · pk )2 + 3. Now let q be an arbitrary prime divisor of N . (Notice that q can’t be 2, since N is 3 mod
4, or 3, since N is 1 mod 3.)Reduce mod q to find that −3 is a square
mod q. So let’s figure out what kinds of primes could have −3 as a
QR. There are two possibilities: either −1 and 3 are both squares, or
they are both nonsquares. If they’re both squares, then we have q ≡ 1
(mod 4), and that means
3
q
=
q
3
So for 3 to be a square mod q, we need q ≡ 1 (mod 3).
Now if both −1 and 3 are nonresidues mod q, then q is 3 mod 4,
meaning that
q 3
=−
q
3
So for 3 to be a nonesidue mod q, we need, once again, q ≡ 1 (mod 3).
In either case, we arrive at the conclusion that any prime divisor of N
must be 1 mod 3. But by construction, as above, any prime congruent
to 1 mod 3 already divides N − 3, and therefore cannot divide N as
well, since clearly it could not divide 3. This contradiction shows that
there are infinitely many primes congruent to 1 mod 3.
3. For this proof, we will make liberal use of the proven properties (not
the definition) of the Jacobi symbol, though we use the definition once,
right now, to claim that n must be odd—otherwise the Jacobi symbol
would be undefined. Now observe that
15
3
5
=
n
n
n
Now, this Jacobi symbol will be 1 in either of two cases—the righthand product is 1 · 1, or else it is −1 · −1.
2
Notice that since 5 ≡ 1 (mod 4), we know that
5
n
=
n
5
So this Jacobi symbol will be 1 when n is either 1 or 4 mod 5, and
−1 when n is 2 or 3 mod 5 (it can’t be 0 by our hypothesis that
(n, 15) = 1).
Now forthe other
Jacobi symbol, there are 2 cases. If n is 1 mod 4,
n
3
then n = 3 , which will be 1 when n is 1 mod 3—and therefore 1
mod 12—and −1 when n is 2 mod 3—and therefore 5 mod 12. On the
other hand, if n is 3 mod 4, these values will be reversed, so the Jacobi
symbol will be 1 when n is 2 mod 3—and therefore 11 mod 12—and
−1 when n is 1 mod 3—and therefore 7 mod 12.
Putting these together, the following cases are those when the Jacobi
symbol is 1:
• When n is both 1 mod 5 and 1 mod 12, or n ≡ 1 (mod 60).
• When n is both 1 mod 5 and 11 mod 12, or n ≡ 11 (mod 60).
• When n is both 4 mod 5 and 1 mod 12, or n ≡ 49 (mod 60).
• When n is both 4 mod 5 and 11 mod 12, or n ≡ 59 (mod 60).
• When n is both 2 mod 5 and 5 mod 12, or n ≡ 17 (mod 60).
• When n is both 2 mod 5 and 7 mod 12, or n ≡ 7 (mod 60).
• When n is both 3 mod 5 and 5 mod 12, or n ≡ 53 (mod 60).
• When n is both 3 mod 5 and 7 mod 12, or n ≡ 43 (mod 60).
These 8 cases are all, by the work above.
4. First we need to know that 15841 = 7 · 31 · 73. This fact will be used
repeatedly.
Let’s show first that 15841 is a Carmichael number. Take an arbitrary
integer a such that (a, 15841) = 1. Note that this clearly implies a
is not divisible by any of the prime factors of 15841. Now we may
compute a15840 mod 7, mod 31, and mod 73. It is easy to check
that 15840 is divisible by 6, by 30, and by 72, so three applications
of Fermat’s Little Theorem tell us that a15840 is 1 mod 7, mod 31,
and mod 73. Hence, by the Chinese Remainder Theorem, we see
that a15840 ≡ 1 (mod 15841), satisfying the definition of a Carmichael
number.
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Now let’s show 15841 is an Euler pseudoprime
to the base 2. We need
2
to compute the Jacobi symbol 15841
, which is the same as
2
2
2
=1
7
31
73
since all of those Legendre symbols are 1 (simply check the three primes
mod 8). So we are done if we can show 27920 ≡ 1 (mod 15841), by
definition of an Euler pseudoprime. This can be done exactly the same
way as the computation above—since 7920 is divisible by 6, 30, and 72,
we may use Fermat’s Little Theorem 3 times followed by the Chinese
Remainder Theorem.
Finally we must show 15841 is an Euler pseudoprime to the base 2. To
do this, we have to write 15840 as a power of 2 times an odd number.
It turns out (and is very easy to check) that 15840 = 25 · 495. So let’s
start by computing 2495 modulo 15481. Remember that if it is either
1 or −1, we are finished. First let’s work mod 7:
2495 ≡ (23 )165 ≡ 1
(mod 7)
2495 ≡ (25 )99 ≡ 1
(mod 31)
2495 ≡ (29 )55 ≡ 1
(mod 73)
Now mod 32,
Finally, mod 73,
(Here I’ve used that 29 = 512 and 73 · 7 = 511, but there are many
other ways to do this computation.) This shows, again by CRT, that
2495 ≡ 1 (mod 495), which suffices to show that 15841 is a strong
pseudoprime to the base 2.
5. Suppose
n is an Euler pseudoprime to the base 3. This means 3(n−1)/2 ≡
3
(mod n). Now, because n is 5 mod 12, n is both 1 mod 4 and 2
n
mod 3, so this means
3
n
2
=
=
= −1
n
3
3
This suffices to satisfy the definition of a strong pseudoprime, as we
have just shown 3(n−1)/2 ≡ −1 (mod n). (Look carefully at the definition.)
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6. We are given that n = 11021 = 103 · 107. Now we may examine this
problem mod 103 and 107 separately, and use CRT at the end. First
let’s look mod 103. We have x4 ≡ 1686 ≡ 38 (mod 103), and we wish
to find x2 . So let’s solve the congruence 4y ≡ 2 (mod 102). Clearly
the value 26 will work for y. Now we have (x4 )26 ≡ x104 ≡ x102 x2 ≡ x2
(mod 103) by Fermat’s Little Theorem, so we need to compute 3826
mod 103. We can do this by successive squaring:
382 ≡ 1444 ≡ 2
4
2
8
2
38 ≡ 2 ≡ 4
38 ≡ 4 ≡ 16
16
38
2
≡ 16 ≡ 256 ≡ 50
(mod 103)
(mod 103)
(mod 103)
(mod 103)
So we have x2 ≡ 3826 ≡ 3816 · 388 · 382 ≡ 50 · 16 · 2 ≡ 1600 ≡ 55
(mod 103).
To look mod 107 is even easier, since here x4 ≡ 1686 ≡ 81 (mod 107).
This will mean (since 107 is prime) that x2 ≡ 9 (mod 107). The
other possibility, −9, cannot be the case because, as 9 is clearly a QR
mod 107, and −1 is not because 107 ≡ 3 (mod 4), −9 is a quadratic
nonresidue mod 107.
So summing up, we have x2 ≡ 55 (mod 103) and x2 ≡ 9 (mod 107);
this implies (by CRT) that x2 ≡ 6750 (mod 11021). (In case you’re
wondering how I solved this, I wrote down the equation 103X + 55 =
107Y + 9, which becomes 107Y − 103X = 46. Then I noticed that
107(26) − 103(27) = 1, so I computed 107 · 26 · 46 + 9 and reduced it
mod 11021.)
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