Mathematics 220
1. Prove that
√
Homework 6
Due Feb. 24/25
6 is irrational.
Solution: Again we make use of the result that “n2 is even iff n is even”.
√
√
Proof. Assume, to the contrary, that 6 ∈ Q. Hence we can write 6 = ab with
a, b ∈ Z. If a, b have a common factors we can always eliminate it and so we can
assume that a, b have no common factors.
2
Now 6 = ab2 and so a2 = 6b2 . Thus a must be even and so we can write a = 2k for
some k ∈ Z. Then 6b2 = 4k 2 and so 3b2 = 2k 2 is even. Claim b2 is even. Assume
to the contrary b2 = 2k + 1 is odd. Then 3b2 = 6k + 3 = 2(3k + 1) + 1 is also odd,
contradicting the above. Therefore b2 is even and so we know b must be even.
Since a and b are both
√ even, they have a common factor of 2. This contradicts our
assumption and so 6 6∈ Q.
2. Prove that
√
2+
√
3 is irrational.
√
√
Solution:√ Assume
contrary that 2 + 3 is rational. Then√there are integers
√ to the
2
a, b s.t.
2 + 3 = ab . Squaring both sides we get 2 + 3 + 2 6 = ab2 , and so
√
√
√
2
6 = 12 ( ab2 − 5) ∈ Q. This contradicts the previous question and so 2 + 3 must
be irrational.
3. Prove that if the integers a and b are both odd, then a2 + b2 cannot be a perfect square.
Solution: Assume to the contrary that there are odd integers whose squares when
summed is a perfect square. Then there are integers a, b and c so that (2a + 1)2 +
(2b + 1)2 = c2 . This implies 4(a2 + a + b2 + b) + 2 = c2 . Therefore c2 is even which
implies c is even and so c = 2d for some integer d. But then 4(a2 +a+b2 +b)+2 = 4d2
and so 4|2 which implies 2/4 is an integer. We have derived a contradiction.
In fact, there is a more elegant solution which is also similar to the solution to the next
problem: fist, prove a lemma, that if a is an integer, then a2 can only be congruent
to 0 or 1 mod 4 (this lemma is proved by cases, in the same way as the lemma that
appears in the solution of the next problem). Now, note that if a is odd, then a2
has to be congruent to 1 mod 4 (prove it!) Thus, if a and b are both odd, then
a2 + b2 ≡ 1 + 1 = 2 mod 4, and thus this number cannot be a square of an integer,
since any square cannot be congruent to 2 mod 4.
4. Prove that the number 123456782 cannot be represented as a2 + 3b2 for any integers a
and b. (Hint: Consider the remainder mod 3).
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Mathematics 220
Homework 6
Due Feb. 24/25
Solution: First, note that 123456782 ≡ 2 mod 3. How did we find out? Of course,
you can just try to find out what the remainder of this number mod 3 is using a
calculator: try subtracting 0, 1 or 2 and dividing by 3. However, a much better way
is by using Problem 5(a) from Workshop 3: this number is congruent to the sum of
its digits mod 3 (make sure you understand why Problem 5(a) from the Workshop
implies this; that problem says something about congruence mod 9). We have:
123456782 ≡ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 2 mod 3, and you do not even have to
evaluate this sum, since we only care about its remainder mod 3, so you can right
away eliminate everything that’s divisible by 3, so 1 + 2 disappears, 3 disappears,
6 disappears, 4 + 5 disappears, 7 gets replaced by 1, and 8 gets replaced by 2, and
finally you get 2 (make sure you understand this calculation!).
Next, observe that for any integers a, b, a2 + 3b2 ≡ a2 mod 3 (Why?).
Finally, let us prove a lemma: if a is an integer, then a2 can be congruent to 1 or 0
mod 3. Proof of this lemma is by cases: the possibilities are, a ≡ 0 mod 3, a ≡ 1
mod 3, a ≡ 2 mod 3. In the first case, a2 ≡ 02 = 0 mod 3, in the second case
a2 ≡ 11 = 1 mod 3; in the third case, a2 ≡ 22 ≡ 1 mod 3, and the lemma is proved.
Now we can complete the proof of our original statement by contradiction. Suppose
there existed a, b such that 123456782 = a2 + 3b2 . Then we would have: 123456782 ≡
a2 mod 3, but by the Lemma, a2 can be congruent only to 0 or 1 mod 3,and our
number is congruent to 2 mod 3, and we arrive at a contradiction.
5. (a) Prove that there are infinitely many primes p such that p ≡ 3 mod 4. Hint: try to
proceed the same way as in Euclid’s proof of the statement that there are infinitely
many prime numbers; but instead of making the number N = p1 , . . . pn + 1, make
a number N that is definitely congruent to 3 modulo 4 (and that still differs by 1
from a number that is divisible by all of p1 , . . . pk ).
(b) Could this proof have worked for the primes congruent to 1 modulo 4?
Solution: Suppose, to the contrary, there were only finitely many prime numbers
of the form 4k + 3. Let us denote them by p1 , . . . , pn . Let N = 4p1 . . . pn − 1. If
N is prime, we get a contradiction, since N is bigger than all the pi , i = 1, . . . , n,
and therefore N is not one of these numbers, on the other hand, N ≡ 3 mod 4, and
thus N would be a prime number of the form 4k + 3 that is not on our list. Then N
has to be a product of prime numbers, by the Fundamental Theorem of Arithmetic
(which we have not yet proved, but we agreed we can use). Note that pi |(N + 1), so
pi cannot divide N, for all i = 1, . . . , n. Note also that N is odd, so 2 is not on the
list of prime factors of N; all the prime numbers that are not equal to 2 are odd, so
they have to be either congruent to 1 or 3 mod 4. Then all the prime factors of N
have to be congruent to 1 mod 4, since we are assuming that the numbers pi are all
the prime numbers that are congruent to 3 mod 4, and none of them can be among
the factors of N. Then we get that N is a product of several factors, each congruent
to 1 mod 4. But then N itself has to be congruent to 1 mod 4 (since we know
that we can multiply congruences) – a contradiction.
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Mathematics 220
Homework 6
Due Feb. 24/25
Part (b). Note that if we tried to make a similar proof for the fact that there are
infinitely many primes congruent to 1 mod 4, it would not work: if we made some
number N, and tried to prove that all its factors cannot be congruent to 3 mod 4,
we would not be able to do that, because the product of two numbers congruent to 3
mod 4 is a number that is congruent to 1 mod 4 (so all we would get is that there
has to be an even number of factors of the form 4k + 3, and we would not be able to
obtain a contradiction with our assumption this way).
6. Find the last digit of the number 20162016 .
Solution: It is 6¿ Indeed, note that the last digit is the remainder of a number
mod 10. We have: 2016 ≡ 6 mod 10, wo 20162016 ≡ 62016 mod 10 by the properties
of congruence proved in class. Now it remains to note that 6n ≡ 6 mod 10 for any n
(note that now that we know mathematical induction, it is the easiest to prove this
statement by induction). At the time the problem was assigned, you could argue, for
example, as follows: 62 ≡ 6 mod 10, then
62016 = (62 )1008 ≡ 61008 ≡ 6504 ≡ 6252 ≡ 6126 ≡ 663
mod 10.
Now, using the same idea, 663 = 6 · 662 ≡ 6 · 631 ≡ 36 · 630 ≡ 6 · 615 mod 10, and so
on – you can keep reducing the power until you just get 6 itself.
7. Are the following statements True or False (provide complete proofs):
(a) If x and y are both irrational, then x + y is irrational.
(b) If x and y are both irrational, then xy is irrational.
(c) If x is rational and y is irrational, then xy is irrational.
(d) If x 6= 0 is rational and y is irrational, then xy is irrational.
√
√
(e) If a, b ∈ Q and ab 6= 0, then a 3 + b 2 is irrational.
Solution: (a) False. Counterexample: x =
√
2, y =
√
2.
(b) False. The same counterexample as in (a) works.
(c) False.
Counterexample: If x = 0 and y is any irrational number, for example,
√
y = 2, xy = 0 is rational.
(d) True. Proof by contradiction. Suppose, to the contrary, that x 6= 0, x = a/b
for some integers a, b, y is irrational, and xy = c/d for some integers c, d. Then
y = xy/x = cb/da is rational as well – a contradiction!
√
√
(e) True. Proof by contradiction. Suppose a 3 + b 2 was rational (and equal to c/d
where c,√d are√integers) for some rational numbers √
a, b with ab 6= 0. Then we would
have (a 3 + b 2)2 = c2 /d2, which implies that 2ab 6 = c2 /d2 − 3a3 − 2b2 . Note that
Page 3 of 4
Mathematics 220
Homework 6
Due Feb. 24/25
the right-hand side is a rational number. Thus we obtained that the product of
√ the
non-zero rational number x = 2ab and the irrational (by Problem 1) number 6 is
rational. This contradicts the statement of Part (d) that we just proved, so we get a
contradiction!
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