An integer $a$ is "happy" if it can be written on the form $a=6k+3$ for some integer k. Let $y$ denote an integer. Prove that $y$ is "happy" if and only if $y^2$ is "happy".
My proof:
Assume that $y$ is a happy integer, that is $y=6k+3$. Then we have $y^2=(6k+3)^2=36k^2+36k+6+3=$ $\color{blue}{\textrm{6($6k^2+6k+1$)}}$$+3$ which is the same as $\color{blue}{\textrm{6$m$}}$$+3$ for some integer $m$. Thus we have that if $y$ is happy then so is $y^2$.
Next we show that there exist no integers that yield a happy square other than those integers who can be expressed on the form $6k+3$.
$1. (6k+1)^2=36 k^2 + 12 k + 1 \neq6m+3$
$2. (6k+2)^2=36 k^2 + 24 k + 4\neq6m+3$
$3. (6k+4)^2=36 k^2 + 48 k + 16 \neq6m+3$
$4. (6k+5)^2=36 k^2 + 60 k + 25 \neq6m+3$
We find that the only way of achieving a happy integer by squaring is by squaring an already happy integer. Since squaring a number is a reversable operation, if we take the square root of a happy number and it is still an integer, then it can only have mapped onto a happy number $6k+3$. Hence we have shown that iff a square is happy then so is its square root (if it is still an integer). ■
My questions:
First of all if it is correct? Furthermore, I feel like the proof have unnecessary steps somehow and would very much like to shorten it, how can one do that?. Also, I have just started to read about how to prove/present the idea of a proof by oneself - was there anything you wished you knew regarding proofs during these early stages that I am going through that might be useful to know before taking on more advance excersices?
How about using prime factorization, as follows. (Just noted this is basically what is suggested in lulu's comment.)
$\Rightarrow$
If $$y = 6k+3 = 3(2k+1),$$ for some $k \in \Bbb Z$, then $y$ must be of the form $$ y = 3\prod_i p_i^{j_i},$$ with $p_i$ odd primes.
So
$$y^2 = 3\cdot\underbrace{\left(3 \prod_i p_i^{2j_i}\right)}_{\small\mbox{odd}},$$
and thus also $y^2$ is of the form
$$y^2 = 3(2h+1),$$ for some $h \in \Bbb Z$
$\Leftarrow$
Conversely, if $z$ is the square of some integer $y$ and
$$z = 3(2k+1),$$ then it must be
$$z = 3\left(3\prod_i p_i^{2j_i}\right),$$ with $p_i$ odd primes, because, the product is odd.
Thus
$$y = 3 \underbrace{\prod_i p_i^{j_i}}_{\small{\mbox{odd}}}.$$ Therefore it must be $$y= 3(2h +1),$$ for some $h\in\Bbb Z$.