1. ## Proving statements

Hi guys, need help with two probs below

1. Prove either true or false:

3|(a^2+b^2) => 3|a and 3|b

2. Prove that prime number 7t+1 can also be written as 14s+1

Much appreciated!

2. ## Re: Proving statements

Hello, lolcat!

1. Prove either true or false:

. . $3|(a^2+b^2) \;\Rightarrow\;3|a\,\text{ and }\,3|b$

This is not a rigorous proof . . .

Note the squares:

. $\begin{array}{cccc} 0^2 &\equiv& 0 & \text{(mod 3)} \\1^2 &\equiv& 1 & \text{(mod 3)} \\ 2^2 &\equiv& 1 & \text{(mod 3)} \\ 3^2 &\equiv& 0 & \text{(mod 3)} \\ 4^2 &\equiv&1 & \text{(mod 3)} \\ 5^2 &\equiv& 1 & \text{(mod 3)} \\ 6^2 &\equiv&0 & \text{(mod 3)} \\ 7^2 &\equiv& 1 & \text{(mod 3)} \\ 8^2 &\equiv&1 & \text{(mod 3)} \\ 9^2 &\equiv& 0 & \text{(mod 3)} \end{array}$

If $a^2+b^2$ is divisible by 3, then: . $a^2+b^2 \:\equiv\:0\text{ (mod 3)}$

We see that the only way that $a^2+b^2 \:\equiv\:0\text{ (mod 3)}$
. . is when both $a$ and $b$ are divisible by 3.

2. Prove that prime number $P \,=\,7t+1$ can also be written as $14s+1.$

If $t$ is odd, $t \,=\,2m-1,$
. . then: . $P \:=\:7(2m-1) + 1 \:=\:14m - 7 + 1 \:=\:14m - 6 \:=\:2(7m -3)$
$P$ is not a prime.

Hence, $t$ must be even, $t = 2s$

Then: . $P \:=\:7(2s) + 1 \:=\:14s + 1$