## 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 $a^p \equiv b^p \mod p.$ $\blacksquare$