Download Solutions to Practice Problems, Math 312 1 Prove that

Solutions to Practice Problems, Math 312
1 Prove that the fourth power of an odd integer is expressible in the
form 16n + 1 for n ∈ Z.
Solution If m is odd, we can write m = 2k + 1 for some k ∈ Z and so
m4 = (2k + 1)4 = 16k 4 + 32k 3 + 24k 2 + 8k + 1
and so we have
4
3
m = 16 k + 2k +
k(3k + 1)
2
+ 1.
It remains to show that k(3k+1)
∈ Z. If k is even, then k2 ∈ Z, while if k
2
is odd, then 3k + 1 is even and so 3k+1
∈ Z. This completes the proof.
2
2 Define an by a0 = 1, a1 = 2, a2 = 4 and
an+2 = an+1 + an + an−1 , for n ≥ 1.
Show that an ≤ 2n for all n ∈ N.
Solution Note the typo - I meant to write a2 = 4. We use induction
on n. The inequality is true for n = 0, 1 and 2. Suppose that it is true
for all n ≤ k where k ≥ 2. Then
ak+1 = ak + ak−1 + ak−2 ≤ 2k + 2k−1 + 2k−2 = 7 · 2k−2 < 2k+1 .
3 If Fn is the nth Fibonacci number, prove that
Fn+1 Fn−1 − Fn2 = (−1)n .
Solution We use Binet’s formula and find that the left-hand-side is
n+1
n−1
n
2
α
− β n+1
α
− β n−1
α − βn
√
√
√
−
5
5
5
and hence is equal to
2(αβ)n − αn+1 β n−1 − αn−1 β n+1
.
5
Since αβ = −1, this is equal to
2 − αβ −1 − α−1 β
2 + α2 + β 2
n
n
(−1)
= (−1)
= (−1)n .
5
5
4 Let a and n be positive integers with a > 1. Prove that, if an + 1
is prime, then a is even and n is a power of 2.
Solution If an + 1 is prime, then, since a > 1, it follows that an + 1
must be odd and so a must be even. If n > 1, then if n is not a power
1
2
of 2, we would have that n is divisible by some odd prime, say p. If
n = p, we have
an + 1 = ap + 1 = (a + 1) ap−1 − ap−2 + · · · − a + 1
and so an + 1 is not prime (each factor above is > 1). If n > p,
an + 1 = (an/p )p + 1 = (an/p + 1) a(n(p−1)/p − an(p−2)/p + · · · − an/p + 1
which again contradicts the fact that an + 1 is prime.
5 Show that if all three of p, p + 2 and p + 4 are prime, then p = 3.
Solution We apply the Division algorithm to p and 3 and find that
p has a remainder of either 0, 1 or 2, after division by 3. In the first
case, we necessarily have p = 3. In the second, it follows that p + 2
is divisible by 3 and hence equal to 3, contradicting the fact that p is
prime. In the third case, we have that p + 4 is divisible by and hence
equal to 3, again a contradiction.
6 Use the Euclidean algorithm to compute (2059, 2581) and to express
this quantity as a linear combination of 2059 and 2581.
Solution We have
2581 = 1 · 2059 + 522
2059 = 3 · 522 + 493
522 = 1 · 493 + 29
493 = 17 · 29 + 0
and so (2059, 2581) = 29. We thus can write
29 = 522 − 493 = 522 − (2059 − 3 · 522) = 4 · 522 − 2059
and so
29 = 4(2581 − 2059) − 2059 = 4 · 2581 − 5 · 2059.
7 Show that every nonzero integer can be uniquely expressed as
ak 3k + ak−1 3k−1 + · · · + a1 3 + a0
where ai ∈ {−1, 0, 1} and ak 6= 0.
Solution This follows from the Division algorithm upon noting that
if, after dividing n by 3, we obtain a remainder of 2, we may write
n = 3k + 2 = 3(k + 1) − 1.
8 Prove that there are infinitely many primes of the shape 6n + 5.
Solution We note that if we multiply 2 numbers of the form 6k + 1
together, we get another number of the same form (check this!). If
3
there were only finitely many primes of the form 6k + 5, say
p0 = 5 < p1 < p2 < · · · < pn .
Consider the integer
N = 6p1 p2 · · · pn + 5.
Clearly N > 1 is not divisible by 2 and the note with which we began
this solution implies that N has at least one prime divisor p of the form
6k + 5. If p = 5, we get that
5|N − 5 = 6p1 p2 · · · pn ,
while, if p > 5, we have that
5|N − 6p1 p2 · · · pn = 5.
In either case, we have a contradiction.