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Find all polynomials $f:R→R$ such that $f(x+2)=f(x)+2$

Find all polynomials $g:R→R$ such that $g(2x)=2g(x)$

Since the functions must be polynomials, I tried using an argument by degrees, but this did not lead me anywhere. Can someone help with some ideas?

I took the degrees of both sides to be d, but this doesn't help since both sides are of the same degree. In my previous question which can be seen on my profile (mentioned by @Scene), multiplying by a polynomial on both sides worked, but I am not able to use this here. –

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    Why has this been downvoted? The OP has formatted things properly and stated what they have tried (albeit not in too much detail). They seem to have posted a different question in the same format with 3 upvotes. Curious as to why this one stands out. – Scene Mar 30 '23 at 05:33
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    What did you do in using the degree? If you use it correctly that would tell you a lot. – Arctic Char Mar 30 '23 at 12:35
  • I took the degrees of both sides to be $d$, but this doesn't help since both sides are of the same degree. In my previous question which can be seen on my profile (mentioned by @Scene), multiplying by a polynomial on both sides worked, but I am not able to use this here. –  Mar 30 '23 at 13:07
  • For the second problem, you can assume leading term of $g(x)$ to be $ax^n, a\neq 0$, then comparing leading coeffs in the equation gives $2^n=2$, hence $n=1$, i.e. $g(x)$ is linear. The only poly that doesn't have non-zero leading coefficient is $g(x)=0$ which can be checked separately. – Sil Apr 02 '23 at 16:23

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For the first one, put $x =0$ to get (Notice that $f(0)$ is the constant term. Call it $c$): $$f(2) = f(0) + 2 = c+2$$ Put $x = 2$ to get $$f(4) = f(2) + 2 = c+2 + 2 = c+4$$ Show by induction on a natural number $n$ that $f(2n) = c+2n$. Define $$h(x):= f(x) - f(0) - x$$ Notice that $h(x)$ has a degree no greater than $f(x)$ (As pointed out by prets, this is true only if $\deg(f)\ge 1$, which is implied by $f(x+2) = f(x)+2$ ). Also, $h(x)$ has infinite roots (all even numbers), so $h(x) = 0$ from here.

For the second question, put $x = 0$ to get: $g(0) = 2g(0) \iff g(0) = 0$. So, we can write: $$g(x) = xp(x)$$ Thus, $$2xp(2x) = 2xp(x)$$ Suppose $x \neq 0$ then we have: $$p(2x) = p(x)$$

It can be shown that $p(x) = p(1)$ is some plain old constant.

D S
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  • Tiny note: $h(x)$ has a degree no greater than $f(x)$ unless $\deg f(x) < 1$, but of course that's impossible because of $f(x + 2) = f(x) + 2$. – Jakob Streipel Mar 30 '23 at 17:33