There are an infinite number of polynomials P for which P(x+5) - P(x) = 2 for all x. What is the least possible value of P(4) - P(2)?

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There are an infinite number of polynomials P for which P(x+5) - P(x) = 2 for all x. What is the least possible value of P(4) - P(2)?

Mathematics
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P(x+5) - P(x) -2 = 0
..
Let \[P(x) = ax^{n}, .... P(x+5) = a(x+5)^{n}\] Then \[a(x+5)^{n} -ax^{n} = 2\] \[\rightarrow a = \frac{2}{(x+5)^{n} -x^{n}}\] since it applies for all x, let x=0 \[a = \frac{2}{5^{n}}\] Evaluate P(4)-P(2) \[\rightarrow \frac{2}{5^{n}}4^{n} - \frac{2}{5^{n}}2^{n} = \frac{2(4^{n}-2^{n})}{5^{n}}\] Take limit as n-> infinity \[\lim_{n \rightarrow \infty}\frac{2(4^{n}-2^{n})}{5^{n}} = 0\] smallest possible value goes to 0

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@dumbcow: why the initial assumption of \( P(x) = ax^{n} \)
\[P(x+5)-P(x)=2\]\[P(x+5)-P(x)-2=0\] if this is the function, given two value of x, 4 and 2, to be subtracted. . \[[P(4+5)-P(4)-2] -[P(2+5)-P(2)-2]=0\]\[[P(9)-P(4)-2]-[P(7)-P(2)-2]=0\]\[P(5)-2 -P(5)-2=0\]\[p(0)-4=0\]\[P(0)=4\] kindly check my solution. i got 0 as the least value of x, and 4 is the result. .
@FoolFormath because it was the simplest polynomial to work with and since it doesn't matter what polynomial it is
@dumbcow, how about my solution? is it acceptable. .
in one of your steps(3rd line) what happened to P(9) and P(7)
oh i got it but how can you assume P(9) -P(4) = P(5) ?
they're functions P
No correct answer yet. Hint: Consider the value(s) of n a polynomial P of degree n would have in order for P(x+5) - P(x) to be constant.
It seems that I have understood the solution (with some help) the polynomial should be for the form \( P(x)=\frac{2}{5}x+c \) so \( P(4)-P(2)=\frac{4}{5} \)
4/5 is correct
i see what i did wrong now. isn't the question a little misleading though, there really is only 1 possible value for P(4)-P(2)

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