## Problem Set 2

PROBLEMS

1.) Find a linear function $f(x)$ such that $f(1) = 42$ and $f(2) = 47$.

2.) Solve for $x$: $4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 170.$

3.) Prove that the product of $3$ consecutive numbers is always divisible by $6$.

4.) Prove that if $p$ is prime, $a$ and $b$ are integers, and $a \equiv b\mod p$, then $a^p \equiv b^p \mod p$.

SOLUTIONS AND PROOFS

Post Date: October 20, 2009

1. Solution: This is just the same as saying, find the equation of the line passing through $(1,42)$ and $(2,1337)$. So, by point slope formula, we have, $y - y_1 = \displaystyle\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\Rightarrow y - 42 = \displaystyle\frac{1337 - 42}{2 - 1}(x - 1). \Rightarrow y = f(x) = -18x + 92.$

2.) Solution:$4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 4^x(1 + 4 +4^2 +4^3) = 170 \Rightarrow 4^x(85) = 170\Rightarrow 4^x = 2 \Rightarrow 2^{2x}=2^1 \Rightarrow x = \displaystyle\frac{1}{2}.$

3.) Proof: A number is divisible by $6$ if it is divisible by $2$ and $3$. A product of $3$ consecutive numbers is divisible by $2$ because at least one of them is even, so it remains to show it is divisible by $3$.

If a number is divided by $3$, its possible remainders are $0, 1,$ and $2$.  Assume $n, n +1$ and $n+2$ be the three consecutive numbers, and $r$ be the remainder if $n$ is divided by $3$.

Case 1: If $r=0$, we are done.

Case 2: If $r = 1$, then $n + 2 \Rightarrow r=0$

Case 3: If $r = 2$, then $n + 1 \Rightarrow r = 0$.

Since the product of the three consecutive numbers is even, and for each case of $r$, one of the consecutive numbers is divisible by $3$, the product of three consecutive numbers is divisible by $6. \blacksquare$

4.) Proof: From definition, $a^p \equiv b^p \mod p \Leftrightarrow b = a + kp$ for some $k \in \mathbb{Z}.$

Raising both sides of the equation to $p$, we have $b^p = (a + kp)^p.$ By the binomial theorem,  $b^p = (a + kp)^p = a^p + \displaystyle {p \choose 1}a^{p-1}kp + \displaystyle{p \choose 2}a^{p-2}k^2p^2 + \ldots + k^pp^p$.

Notice that every term aside from $a^p$ is divisible by $p^2$. (Why?). Therefore,  $a^p \equiv 0 \mod p^2 .$

Hence, then $a^p \equiv b^p \mod p.$ $\blacksquare$

## Problem Set 1

PROBLEMS

1.) The sum of two numbers is $18$ and there difference is $-4$. What are the two numbers?

2.) Find the values of $p, q$ and $r$ if:

$p + q + r = 3 \frac{1}{2}$
$pq + qr + rp = - 2 \frac{1}{2}$
$pqr = -2$

3.) Prove that $\displaystyle\frac{x + y}{2} \geq \sqrt{xy}$

4.) Define $cos(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$ and $sin(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$

a.) Prove that $cos(-x) = cos(x)$
b.) Prove that $sin(x) = -sin(x)$

SOLUTIONS AND PROOFS
Posted October 13, 2009

1.) Solution: Let $x$ and $y$ be the two numbers. Then, $x + y = 18$ and $x - y = -4$. Adding the equations, we have $(x + y) + (x - y) = 18 + (-4) \Rightarrow 2x = 16 \Rightarrow x = 8$. Substituting it to the first equation gives us $y = 10$. Therefore, the two numbers are $6$ and $10$.

2.) Solution: From the given, $p, q$ and $r$ are roots of of the cubic equation $x^3 - \frac{7x^2}{2} - \frac{5x}{2} + 2.$ Factoring, we have $(x+1)(x-4)(2x + 1)=0.$ Therefore, $x = 1, 4$ or $-1/2$

3.) Proof: We know that the square of the difference of any two numbers is always positive or $0$. Let $x, y$ be any two numbers. Then, $(x - y)^2 \geq 0$. Expanding, we have $x^2 - 2xy + y^2 \geq 0$. Adding $4xy$ to both sides of the equation yields $x^2 + 2xy + y^2 \geq 4xy \Rightarrow (x + y)^2 \geq 4xy$. Getting the square root of both, we have, $x + y \geq 2 \sqrt{xy} \Rightarrow \frac{x + y}{2} \geq \sqrt{xy}. \blacksquare$

4.) Proof (a): We want $\cos (-x)$ so we will just replace $x$‘s with $-x$. Therefore, $cos(-x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-x)^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-1 \cdot x)^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-1)^{2n} \cdot x^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}= cos(x). \blacksquare$

Proof of 4b is left as an exercise. It’s very similar to the proof of 4a.

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