Question

prove that if n is an integer and 3n+2 is even then n is even using

Answer 1

odd×odd=oddodd×odd=odd
odd+even=oddodd+even=odd

Thus, if nn was odd, then:

3n+2=odd×odd+even=odd+even=odd3n+2=odd×odd+even=odd+even=odd

But we know that 3n+23n+2 is even - thus, nncan't be odd.






Another way of doing it: Say nn is odd. Thus, there exists a kk such that n=2k+1n=2k+1. Then:

3n+2=n=2k+13(2k+1)+2=6k+3+2=6k+5=6k+4+1=2(3k+2)+13n+2=n=2k+13(2k+1)+2=6k+3+2=6k+5=6k+4+1=2(3k+2)+1

Anything of the form 2j+12j+1 is odd. (In this case, j=3k+2j=3k+2.) Thus, we have that 3n+23n+2is odd.

But we know that 3n+23n+2 is even - thus, nncan't be odd.