To prove that [tex]\( n^2 - 2 - (n - 2)^2 \)[/tex] is always an even number for any integer [tex]\( n > 1 \)[/tex], we'll simplify the expression step by step.
Given expression:
[tex]\[ n^2 - 2 - (n - 2)^2 \][/tex]
Let's first expand [tex]\( (n - 2)^2 \)[/tex]:
[tex]\[ (n - 2)^2 = n^2 - 4n + 4 \][/tex]
Now, substitute this expanded form back into the original expression:
[tex]\[ n^2 - 2 - (n^2 - 4n + 4) \][/tex]
Next, distribute the negative sign and combine like terms:
[tex]\[ n^2 - 2 - n^2 + 4n - 4 \][/tex]
Combine [tex]\( n^2 \)[/tex] and [tex]\( -n^2 \)[/tex]:
[tex]\[ 0 - 2 + 4n - 4 \][/tex]
Combine the constant terms:
[tex]\[ 4n - 6 \][/tex]
Now we have the simplified expression:
[tex]\[ 4n - 6 \][/tex]
To determine if this expression is always even, let's factor out a common factor:
[tex]\[ 4n - 6 = 2(2n - 3) \][/tex]
Since [tex]\( 2n - 3 \)[/tex] is an integer (because [tex]\( n \)[/tex] is an integer), multiplying it by 2 will always yield an even number. Therefore:
[tex]\[ n^2 - 2 - (n - 2)^2 = 2(2n - 3) \][/tex]
Thus, [tex]\( n^2 - 2 - (n - 2)^2 \)[/tex] is always even for any integer [tex]\( n > 1 \)[/tex].
