Problem Solving Answers to Problem Set 9

12 July 2012

1. Let x, y, and z be integers such that n = x^{4} + y^{4} + z^{4} is divisible by 29. Show that n is divisible by 29^{4}. Answer: Hopefully we shall find that x, y, z are all divisible by 29. (In fact, this must be the case if the result is true. For suppose there was a solution x, y, z with x, say, not divisible by 29. Then x + 29, y, z would satisfy the first condition, but not the second since (x + 29)^{4} − x^{4} is not divisible by 29^{4}. Consider the multiplicative group (ℤ/29)^{∗} formed by the non=zero residues 1, 2, …, 28 mod 29. We know that this group is cyclic (the Primitive Root Theorem), isomorphic to C_{28}. It follows that the 4th powers modulo 29 form a cyclic group of order 7. (If π is a primitive root modulo 29 then these are the elements π^{4i} for 0 ≤ i < 7. This means there are only 7 4th powers we need consider. We have

1^{4} ≡ 1 mod 29,
2^{4} ≡ 16 ≡ −13 mod 29,
3^{4} ≡ 81 ≡ −6 mod 29,
4^{4} ≡ (2^{4})^{2} ≡ 13^{2} = 169 ≡ 24 ≡ −5 mod 29,
6^{4} ≡ 2^{4} · 3^{4} ≡ 13 · 6 = 78 ≡ 20 ≡ −9 mod 29,
8^{4} ≡ (4^{4})^{2} ≡ 25 ≡ −4 mod 29,
9^{4} ≡ (3^{4})^{2} ≡ 6^{2} ≡ 7.

So the 4th powers are 1, 7, −4, −5, −6, −9, −13. It is clear that no three of these numbers can add up to +29 or −29. So if they add to 0 mod 29 they will have to actually add to 0. So there will have to be one or two positive numbers. There cannot be two positive numbers, since −2, −8, −14 are not in the list. So there must be one positive number and two negative numbers. But again, we cannot get −1 or −7 as a sum of two of the negative numbers. Also, we cannot have just one of x, y, z divisible by 29 since the remaining two numbers would have to be ≡ ±n mod 29, and there is no such possibility. We conclude that x, y, z are all divisible by 29, and the result follows.

2. Find all continuous odd functions f : ℝ → ℝ such that the identity f(f(x)) = x holds for all real x. Answer: The identity I(x) = x and J(x) = −x are two such functions. But are they the only ones? Evidently

f(0) = 0

since

f(0) = f(−0) = −f(0).

Also f is bijective, being its own inverse. Since it is continuous, it is a homeomorphism. It follows that the two components of ℝ \ {0}, namely

ℝ^{+} = {x ∈ ℝ : x > 0} and ℝ^{−} = {x ∈ ℝ : x < 0}

must map into one another. If f swaps the two components, then Jf (which must also be a solution to the problem) maps each component into itself. We may assume therefore that this is true of f. A homeomorphism of ℝ is necessarily strictly monotone. In this case since f(0) = 0 it must be monotone increasing.

Suppose f is not the identity, say

f(x) = y > x.

Applying f,

f(f(x)) = x > f(x),

which is a contradiction. Similarly if f(x) < x. Hence f = I, and I, J are the only functions satisfying the conditions in the question.