Find the function $f(x)$. Let $f(x)$ be a polynomial function. If $f(x+2) - f(x) = 8x - 2$ and $f(0) = 5$, then what is $f(x)$?
I tried to replace $x$ with $0,2,4,\ldots $ for discovering some regular pattern but I have no idea after doing that.

Let's try to determine the degree of the polynomial $f(x).$
If $\ f(x)\ $ has degree $0$, then $f(x) = a_0\ $ which implies $\ f(x+2)-f(x) = 0 \neq 8x-2.$
If $\ f(x)\ $ has degree $1$, then $f(x) = a_0 + a_1x\ $ which implies $\ f(x+2)-f(x) = 2a_1 \neq 8x-2.$
If $\ f(x)\ $ has degree $2$, then $f(x) = a_0 + a_1x + a_2x^2\ $ which, after some cancelling, implies $\ f(x+2)-f(x) = 4a_2x + (2a_1+4a_2).$ For this to equal $8x-2,\ $ we must have $a_2 = 2\ $ and $a_1 = -5.$ So $f(x) = a_0 - 5x + 2x^2.\ $ The last condition, $f(0) = 5\ $ implies that $a_0 = 5.$ So $f(x) = 5 - 5x + 2x^2\ $ is certainly a solution - the only solution of degree 2, since we got there by deduction.
Are there any other solutions (of degree $\geq 3$)?
Well if the degree of $f(x) \geq 3,\ $ then $$f(x+2)-f(x) = a_0 + a_1(x+2) + a_2(x+2)^2 + ... + a_n(x+2)^n - (a_0 + a_1x+a_2x^2+...+a_nx^n),\ $$ where $n\geq3.\ $ From the binomial expansion, the $x^{n-1}\ $ coefficient of $f(x+2)-f(x)\ $ is equal to $a_{n-1} + 2na_n - a_{n-1} = 2na_n.\ a_n \neq 0\ $ and $n \neq 0,\ $ therefore the $x^{n-1}\ $ term in $f(x+2)-f(x)\ $ has at degree of at least $2\ $, and so $f(x+2)-f(x)\neq 8x-2.$ So no, there are no other solutions.
