Let n be any even number
Let m be any other even number
(Note m could take on the same value as n)
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Then if xx is odd we can write x -> n+1x→n+1
Then (n+1)^2 -> n^2+2n+1(n+1)2→n2+2n+1
Then (n+1)^3 -> n^3 +3n^2 +3n +1(n+1)3→n3+3n2+3n+1
There will always be a 1 times 11×1 which when added to all the even numbers (n's) makes the total odd.
If xx is divisible by 2 then it is even and x^6x6 will also be even.
color(blue)( x^6 " can be either even or odd")x6 can be either even or odd
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If x^2x2 is odd then let it be represented by (n+1)(n+1)
We have 2x^22x2 which is (n+1) + (n+1)(n+1)+(n+1) which becomes 2n+22n+2 which is even.
color(blue)(2x^2 " is always even")2x2 is always even
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color(blue)(3 " is always odd")3 is always odd
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So it boils down to the behaviour of combining 3 and x^6x6
Let x^6 -> (n+1) -> "odd"x6→(n+1)→odd
Let 3 -> (m+1) -> "odd"3→(m+1)→odd
color(red)("Then "x^6 + 3 -> (n+1) + (m+1) = m+n+2" which is even")Then x6+3→(n+1)+(m+1)=m+n+2 which is even
Likewise if n is even color(red)("then "x^6 + 3 -> (n) + (m+1) = m+n+1" which is odd")then x6+3→(n)+(m+1)=m+n+1 which is odd
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Conclusion:
f(x)f(x) can be either of odd or even.
