I'm stuck on my proof of this concept and I could use some help on understanding what to do next.
Prove there do not exist natural numbers $m$ and $n$ such that $7m^2=n^2$
Proof will be by contradiction. So, assume there does exist natural numbers such that $7m^2=n^2$ is true.
Let $m=p_1^{r_1} p_2^{r_2 }…p_k^{r_k}$ and $n=p_1^{t_1} p_2^{t_2}…p_s^{t_s}$
Then $m^2=(p_1^{r_1} p_2^{r_2 }…p_k^{2r_k})^2=p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k}$ and $n^2=(p_1^{t_1} p_2^{t_2}…p_s^{t_s})^2=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$
Substitute this into $7m^2=n^2$ to obtain,
$7(p_1^{2r_1} p_2^{2r_2 }…p_k^{2r_k})=p_1^{2t_1} p_2^{2t_2}…p_s^{2t_s}$
The idea is that all values of m and n chosen to satisfy $7m^2=n^2$ are natural numbers so they therefore can be expressed as a product of primes.
This is where I'm stuck. How do I go about showing the contradiction?
Thanks for the help
Best Answer
Your indexes here are the same implying that $n$ and $m$ have the same prime factors.
Better to write this as $m=p_{a_1}^{r_1} p_{a_2}^{r_2 }…p_{a_k}^{r_k}$ and $n=p_{b_1}^{t_1} p_{b_2}^{t_2 }…p_{b_s}^{t_s}$.
Okay that means that there is a $7$ on the LHS so there must be a $7$ on the RHS. So one of the $p_{b_i}$ must be equal to $7$. Without loss of generality let's assume that $p_{b_1} = 7$ then we have
Now $2t_1 \ne 1$ and $2t_1 > 1$ so that $7$ on the LHS can't be the only $7$ in the factorization of $7m^2$. So one of the $p_{a_1}$ must be $7$. Without loss of generality let's assume that $p_{b_1} = 7$. then we have
$7m^2 = 7\cdot 7^{2r_1} p_{2a_2}^{2r_2 }…p_{a_k}^{2r_k}=7^{2r_1 + 1} p_{2a_2}^{2r_2 }…p_{a_k}^{2r_k}=n^2 = 7^{2t_1} p_{2b_2}^{2t_2 }…p_{b_s}^{2t_s}$
But that means $2r_1 + 1 = 2t_1$. But that's impossible as $2r_1 + 1$ is odd and $2t_1$ is even.