Final Practice problems : Solutions
1. Note that a3 ≡ 0, ±1
that, if 3 6 | xyz, we have
(mod 9), for any integer a (via Euler’s theorem), so
x3 + y 3 ≡ ±2
(mod 9),
contradicting the fact that x3 + y 3 = z 3 .
2. We have that
while
77
23
77
59
=
=
8
23
18
59
=
=
2
23
2
59
=1
= −1,
so that the congruence x2 ≡ 77 (mod 23 · 59) has no solutions. On the other
hand,
26
3
2
=
=−
=1
23
23
3
while
2
13
13
7
13
−1
26
=
=−
=−
=−
=−
= 1,
59
59
59
59
13
7
7
so that, via the Chinese Remainder Theorem, there are 2 × 2 = 4 solutions to
the corresponding congruence.
3. We want to find integers such that
(y + 1)2 + y 2 = z 2 ,
i.e.
(2y + 1)2 − 2z 2 = −1.
Notice that
(xk +
√
2yk )(1 +
√
2) = (xk + 2yk ) +
√
2(xk + yk ).
We thus have that
(xk +2yk )2 −2(xk +yk )2 = x2k +4xk yk +4y+k 2 −2x2k −4xk yk −2yk2 = 2yk2 −x2k = −1.
We conclude by choosing, for any k, z = xk + yk and
y = yk +
xk − 1
.
2
4. Consider the equation modulo 9, using that a3 ≡ 0, ±1 (mod 9), for any
integer a. If 3 does not divide xyz, then x2 + 2y 3 + 4z 3 cannot be congruent
1
to 0 modulo 9 and hence 3 must divide one of x, y and z. Continuing with
tghis argument, we may show that, in fact 3 | x, 3 | y and 3 | z, so that 3 | w.
Dividing the equation by 27, we find a “smaller” solution :
(x/3)3 + 2(y/3)3 + 4(z/3)3 = 9(w/3)3 .
We conclude by descent.
5. We have, by quadratic reciprocity,
p
5
.
=
p
5
It follows that this equals 1 precisely when p ≡ ±1
(mod 10).
(mod 5), i.e. when p ≡ 1, 9
6. We have
637 = 72 · 13 = 72 (22 + 32 ) = 142 + 212
and
20022002 = 20021001
2
+ 02 .
For 6!, the only problem prime is 3, which divides it to an even power (so
that 6! is the sum of two squares). For 10!, the prime 7 divides it once, so that
10! is not the sum of two squares.
7. The key idea is that for large enough j, there is always a prime p ≡ 3
(mod 4), with j < p < 2j. Applying this with k = 2j or k = 2j − 1, it follows
that k! is divisible by exactly one coppy of p. It is not trivial (but not too hard)
to show that we can in fact find such a prime for j ≥ 4 and can check that this
implies that the largest k for which k! is the sum of two squares is k = 6.
8. This is just the definitions of the Legendre symbol and of congruence.
9. We have
√
91 = [9; 1, 1, . . .],
and
92 − 91 · 12 = −10,
102 − 91 · 12 = 9,
so that
gcd(10 − 3, 91) = 7
is a factor of 91. We thus have 91 = 7 · 13. For 85, we argue similarly.
10. We have
−3
p
=
−1
p
2
3
.
p
Applying what we know about
−1
p
−3
p
and quadratic reciprocity, it follows that
=
p
3
in all cases, which gives the desired result.
11. If n ≡ 5
we have
(mod 12), then we can write n = 12k + 5 for some integer k, and
3
12k + 5
2
=
=
= −1.
12k + 5
3
3
Since n is a base 3 Euler pseudoprime, we thus have
3
3(n−1)/2 ≡
≡ −1 (mod n)
n
and so n is a strong base 3 pseudoprime.
12. We have
hp
i
p
n2 − 2, a0 =
n2 − 2 = n − 1,
√
1
n2 − 2 + n − 1
α1 =
, a1 = 1,
=
α0 − a0
2n − 3
√
n2 − 2 + n − 2
1
α2 =
, a2 = n − 2,
=
α1 − a1
2
√
1
n2 − 2 + n − 2
=
, a3 = 1,
α3 =
α2 − a2
2n − 3
p
1
α4 =
= n2 − 2 + n − 1, a4 = 2n − 2,
α3 − a3
α0 =
and
α5 =
so that
p
1
= α1 ,
α4 − a4
n2 − 2 = [n − 1; 1, n − 2, 1, 2n − 2].
3