If x is an element of the integers and x^2 is even, then x is even.

assume x is odd when x satisfies the hypothesis. When you square an odd you get an odd because odd times odd is odd. so x^2 is odd and x^2 is even which is falser because a number cannot be even and odd

im unsure if this is right

2. sorry i forgot i had one more question.

if a,b are elements of the integers and ab is odd, then a and b are both odd.

assume for sake of contradiction a and b are not both odd when the hypothesis is satisfied. so a could be even and b odd. an even number times an odd is even so ab is even. ab cannot be even and odd so there cannot be a counterexmple

sorry i forgot i had one more question.

if a,b are elements of the integers and ab is odd, then a and b are both odd.

assume for sake of contradiction a and b are not both odd when the hypothesis is satisfied. so a could be even and b odd. an even number times an odd is even so ab is even. ab cannot be even and odd so there cannot be a counterexmple
Claim 1. Let $a,b$ be odd, then $a*b$ is odd.
Proof of Claim 1. Assume the contrary that $a,b$ are not both odd but $a*b$ is odd. Then there are two possible cases as follows.
(i) $a$ is odd and $b$ is even. Then for some $n,m\in\mathbb{Z}$, we may write $a=2*n-1$ and $b=2*m$, which yields $a*b=(2*n-1)*(2*m)=2*(2*m*n)-2*m=2*(m*(2*n-1))$ (even). Hence this case is not possible.
(ii) $a$ and $b$ are both even. Then for some $n,m\in\mathbb{Z}$, we may write $a=2*n$ and $b=2*m$, which yields $a*b=(2*n)*(2*m)=2*(2*m*n)$ (even). Hence this case is not possible either.
Therefore, if $a,b$ are odd, then $a*b$ is odd. $\rule{0.2cm}{0.2cm}$

Claim 2. If $x^{2}$ is even and $x$ is an integer, then $x$ is even.
Proof of Claim 2. Again assume the contrary that $x^{2}$ is even but $x$ is odd. Then, we have from Claim 1 that $x^{2}=x*x$ is odd.
This contradiction proves Claim 2. $\rule{0.2cm}{0.2cm}$