Question

Prove the statement:
For all integers n, if n is odd, then n2 - 3 is even.

Answer 1

any odd number doubled is even;
a even minus a odd is odd.
No this statement is NOT TRUE.

Answer 2

I think it's squared, @rrrrr
\[n^2 - 3\]

Answer 3

oh
ok
then any odd number squared is odd proved here by the units digit:
1^2=1
3^2=9
5^2=25
7^2=49
9^2=81
all of them are odd and you can add tens digit by adding (n0)^2 where n=the tens digit of the number
same with the hundreds digit
And then a odd minus a odd is even, based on that 1-1=0 mod 2.

Answer 4

@rrrrr I'd have approached it this way...
If n is odd, then n = 2k+1, where k is an integer....
\[n^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1\]
Thus proving that \(n^2 \) is also odd. Anyway, good job.