Download Solutions to selected homework problems

Solutions to selected homework problems
Kiumars Kaveh
October 8, 2011
Problem: Find and prove a formula for the sum of first n Fibonacci numbers
with even indices, i.e., f2 + f4 + · · · + f2n .
Solution: By looking at the first few Fibonacci numbers one conjectures that
f2 + f4 + · · · + f2n = f2n+1 − 1.
We prove this by induction on n. The base case is 1 = f2 = f3 − 1 = 2 − 1.
Now suppose the claim is true for n − 1, i.e., f2 + f4 + · · · + f2n−2 = f2n−1 − 1.
Then:
f2 + f4 + · · · + f2n−2 + f2n
= f2n−1 + f2n − 1,
= f2n+1 − 1,
as required.
√
√
√
Problem: Let α = (1 + 5)/2 and β = (1 − 5)/2. If an = (1/ 5)(αn − β n )
show that an = fn .
Solution:
One know√that α and β are solutions of√the equation x2 = x + 1. We verify
that a1 = (1/ 5)(α − β) = 1 and a2 = (1/ 5)(α2 − β 2 ) = 1. Next,
√
an−2 + an−1 = (1/√5)((αn−2 − β n−2 ) + (αn−1 − β n−1 )),
= (1/√5)(αn−2 (α + 1) − β n−2 (β + 1)),
= (1/ 5)(αn−2 α2 − β n−2 β 2 ),
= an .
This shows that an = an−1 + an−2 and hence an = fn for any n.
Problem: Show by induction that for any integer n > 0 we have 5|n5 − n.
Solution: The statement is clearly true for n = 1. Now suppose the statement is true for n, we show that it is true for n + 1. By the bionomial theorem
we have:
(n + 1)5 − (n + 1)
=
n5 + 5n4 + 10n3 + 10n2 + 5n + 1 − n − 1,
1
= n5 + 5n4 + 10n3 + 10n2 + 5n − n,
= n5 − n + 5(n4 + 2n3 + 2n2 + n),
which is divisible by 5 since n5 − n is divisible by 5 (by induction hypothesis).
Problem: Show that every nonzero integer can be uniquely represented as:
ek 3k + ek−1 3k−1 + · · · + e1 3 + e0 ,
where ej = −1, 0, 1 and ek 6= 0.
Solution: To prove that any number can be represented this way just mimic
the proof of Theorem 2.1. For the uniqueness suppose: ek 3k + ek−1 3k−1 + · · · +
e1 3 + e0 = e0k 3k + e0k−1 3k−1 + · · · + e01 3 + e00 . Then (ek − e0k )3k + · · · + (e1 − e01 )3 +
(e0 − e00 ) = 0. Note that for all j, −3 < |ej − e0j | < 3. Since the right-hand side
is divisible by 3 we conclude that e0 − e00 = 0 i.e. e0 = e00 . Canceling a 3 from
both side we obtain: (ek − e0k )3k−1 + · · · + (e2 − e02 )3 + (e1 − e01 ). Repeating the
same argument we get e1 = e01 . Continuing we see ej = e0j for all j.
Pn
Problem: Show that, for any m, j=1 j m is O(nm+1 ).
Pn
Pn
Solution: For all j = 1, . . . , n we have j m ≤ nm . Thus j=1 j m ≤ j=1 nm =
n · nm = nm+1 .
Problem: Use Bertrand’s principle to show that every integer x ≥ 7 can be
written as a sum of distinct primes (more than one prime).
Solution: By Bertrand’s principle, for any k, there is a prime p with k < p <
2k. It follows that for any x (odd or even) there is a prime p with [x/2] < p < x.
We prove the statement by (strong) induction. First 7 = 2+5. Now suppose every integer less than y with 7 ≤ y < x can be written as a sum of distinct primes.
We would like to show that x can be writen as a sum of distinct primes. By the
above we can find a prime p with [x/2] < p < x. Now consider y = x − p. Since
p > [x/2] we see that p ≥ x/2 and hence 2p ≥ x which implies that p ≥ x−p = y.
Now by induction hypothesis we can write y = p1 + · · · + pr where the pi are
distinct primes and all of them are less than y. It follows that none of these
primes is equal to p ≥ y and hence x = p+p1 +· · ·+pr is a sum of distinct primes.
Problem: Use Euclid’s proof for infinitude of prime numbers to show that the
n−1
n-prime pn is no larger than 2(2 ) .
Solution: We prove the statement by induction. For n = 1 we have 21 =
k−1
(20 )
2 . Next suppose the statement is true for any k < n, i.e. pk ≤ 2(2 ) . By
Euclid’s proof we know that pn ≤ (p1 p2 · · · pn−1 ) + 1. Thus:
pn
≤
≤
≤
≤
0
1
2
n−2
(2(2 ) · 2(2 ) · 2(2 ) · · · 2(2
n−2
2(1+2+···+2 ) + 1,
n−1
2(2 −1) + 1,
n−1
2(2 ) ,
2
)
) + 1,
which proves the claim.
Problem: Show that (fn+2 , fn ) = 1, where fn denotes the n-th Fibonacci
number.
Solution: See the text book (Section on Euclidean algorithm) for the proof
of (fn+1 , fn ) = 1 (proof by induction). Now, (fn+2 , fn ) = (fn+1 + fn , fn ) =
(fn+1 + fn − fn , fn ) = (fn+1 , fn ) = 1.
Problem: Show that the number of times n! is divisible by a prime p is equal
to:
[n/p] + [n/p2 ] + [n/p3 ] + · · · .
Solution: Note that eventually pk > n and hence [n/pk ] = 0, that is, the
above sum is a finite sum (after a while all the terms are zero). To prove the
statement notice that the number of multiples of an integer m > 0 between 1
and n is equal to [n/m]. Now any multiple of p between 1 and n gives rise
to (at least) one factor of p in n! = 1 × 2 × · · · × n. Also each multiple of p2
gives rise to (at least) two factors of p in n! and so on. This proves the statement.
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