>>16991394
f(x) = x^4 + 2 x^2 + 1
f(5) = 676

Let f(x) = g(x) + 1, since f(0) = 1

Write g(x) as a power series. Then for all x,

g(2x^2) + g(x) * g(2x^2) = g(2x^3)

But you can show by explicitly writing the power series that this is equivalent to

g(x^2) + g(x) * g(x^2) = g(x^3)

It's trivial to check that g(1) = 0, so that f(1) = 1

f(1) * f(2) = f(3), and so f(2) = f(3) = 62.5

It's also trivial to check that f(x) is an even function [the exercise is left to the reader]

>>16991394
>the coefficients are obviously integer
>problem says real coefficients
why are mathfags like this?

>>16991438
Yep, nice.

Ignoring the f(2) + f(3) = 125 restriction does the functionial equation have any solutions other than the trival
[eqn] f(x) = \sum_{k=0}^n {n \choose k} x^{2k} [/eqn]
ones?

>>16991438
how did you do it ?
>inb4 AI

>>16991458
NTA but my guess is that you can breakdown the initial condition f(2)+f(3)=125=25+100, which is just the sum of two squares, so f(2)=25=5^2 and f(3)=100=10^2. From there you can find the appropriate polynomial using f(0)=1.

Another thing to notice is that f(2)=5^2=(1+4)^2, f(3)=100=(1+9)^2, ... , f(n)=(n^2+1)^2. So f(5)=(5^2+1)^2=26^2=676.

>>16991454
Doesn't seem like it
Let the leading coefficient and degree of [math]f(z)[/math] be [math]a[/math] and [math]d[/math] respectively, then the leading coefficients of the sides of the functional equation are:
LHS:[math]a^2 2^{d}\quad[/math]RHS: [math]a2^{d}\quad a^2 2^{d}=a2^{d}\rightarrow a=1[/math] ([math]f(0)\neq0[/math] so [math]a\neq0[/math]), this implies [math]f(z)=1[/math] in the constant case
Otherwise, let [math]z[/math] be a root of [math]f(z)[/math], suppose for contradiction, we have [math]|z|>1[/math], note if [math]z[/math] is a root, [math]g(z)=2z^3+z[/math] is a root
[math]|2z^3+z|\ge||2z^3|-|z||[/math] by the reverse triangle inequality, and since [math] |z|>1\rightarrow |2z^3|>2|z|,\quad||2z^3|-|z||>|2|z|-|z||=|z|[/math]
Because [math]|2z^3+z|>|z|[/math] we can construct infinite distinct roots by iterating [math]g[/math], contradicting [math]f(z)[/math] being polynomial, so [math]|z|\leq1[/math] for all roots [math]z[/math]
As [math]a=1,\;f(0)=1\rightarrow |\prod_{f(z)=0}z|=1[/math], since [math]|z|\le1[/math], this requires [math]|z|=1[/math], and therefore [math]|2z^3+z|=|z|\rightarrow|2z^2+1|=1\rightarrow z=\pm i[/math]
So if a root exists, it must be [math]\pm i[/math], the only variable is multiplicity. In the constant case, we set [math]d=0[/math]
[math]f(z)[/math] has real coefficients, so [math]i[/math] and [math]-i[/math] are roots of the same multiplicity [math]m[/math] where [math]2m=d[/math]
Since [math]a=1[/math], the solutions to the functional equation are:
[eqn]f(z)=(z+i)^m(z-i)^m=(1+z^2)^m=\sum_{k=0}^m \binom{m}{k}z^{2k}\qquad m\ge0[/eqn]

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>>16991394

f(0) = 1^2
f(1) = 2^2
f(2) = 5^2
f(3) = 10^2

f(x) = (x^2 + 1)^2

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>>16991921

f = y^2

y(0) = 1
y(1) = 2
y(2) = 5
y(3) = a = 10

y(1) – y(0) = 2 – 1 = 1
y(2) – y(1) = 5 – 2 = 3
y(3) – y(2) = a – 5 = 5

If Δy is linear (and it appears to be),
then y is quadratic.

And solving the depicted equation for y,
yields y = x^2 + 1.

>>16991921

f(2) + f(3) = 125 [is given]
and
f(1)*f(2) = f(3) [is due to the product condition]
thus
f(2) + f(1)*f(2) = 125
or
[1 + f(1)]*f(2) = 125

This results in four cases:
1 + f(1) = 1, f(2) = 125
1 + f(1) = 5, f(2) = 25
1 + f(1) = 25, f(2) = 5
1 + f(1) = 125, f(2) = 1

In other words:
f(0) = 1, f(1) = 000, f(2) = 125, f(3) = 000
f(0) = 1, f(1) = 004, f(2) = 025, f(3) = 100
f(0) = 1, f(1) = 024, f(2) = 005, f(3) = 120
f(0) = 1, f(1) = 124, f(2) = 001, f(3) = 124

f oscillates in cases 1, 3, and 4.
And f increases in case 2.
Considering that this is /sci/,
case 2 must be the case.

>>16991458
>how did you do it ?
Thanks for asking.
Here's exactly how I did it:
I did this 1st: >>16991934
I did this 2nd: >>16991930
I did this 3rd: >>16991921
Thus I posted things in reverse chronological order.

>>16991934
>This results in four cases:
>1 + f(1) = 1, f(2) = 125
>1 + f(1) = 5, f(2) = 25
>1 + f(1) = 25, f(2) = 5
>1 + f(1) = 125, f(2) = 1
this assumes integer coefficients. the coefficients are real. 125 can factor into 62.5 and 2, for example.
>case 2 must be the case.
does not follow from the conditions of the problem

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>>16991943
You don't contradict or dispute me like that.
I only want to be surrounded by "yes"-men.
I don't want to be surrounded by any "no"-persons.
Is that clear?

f(0) = A = 1
f(1) = B = 125/C – 1
f(2) = C
f(3) = D = 125 – C

A < B < C < D
[or f is locally increasing]
implies
3*√14 – 1 < C < 125/3

1*√A – 3*√B + 3*√C – 1*√D = 0
[or y = √f is quadratic]
implies
C = 25

>>16991990
> Is that clear?
No. Please elaborate more.

676

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>>16992020
Ayo it's palindromic six-seven lmfao

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doin it the hard way because fuck you

>>16992114
Sorry, chud, you assumed a_i is positive for all a. This need not be the case.

>>16991394
math has numbers not letters and )*#(&) shit. Nobody actually does any of that, its just shit you put on a math test to waste kids time.

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>>16992114
These 'e-2's were supposed to be just e.

>>16992215
I didn't tho