Extracurricular Elementary Mathematics Marathon (1 Viewer)

KingOfActing

lukewarm mess
1. Does there exist a non-constant polynomial P(x) with integer coefficients such that P(k) is prime for every positive integer k?

I'm not sure if this works but:

Suppose there exists such a polynomial. P(0) =/= 0, as otherwise the x | P(x). Suppose P(0) = p for some prime number p. Then P(p) = ap^n + bp^n-1 + ... + p which is divisible by p, and is such a composite number. (1 is not a prime number, right?)

seanieg89

Well-Known Member
I'm not sure if this works but:

Suppose there exists such a polynomial. P(0) =/= 0, as otherwise the x | P(x). Suppose P(0) = p for some prime number p. Then P(p) = ap^n + bp^n-1 + ... + p which is divisible by p, and is such a composite number. (1 is not a prime number, right?)
Being divisible by p does not imply that you are a composite number. Why can't P(p)=p itself?

KingOfActing

lukewarm mess
Being divisible by p does not imply that you are a composite number. Why can't P(p)=p itself?
I think I'm grasping at straws here, but:

P(kp) must be divisble by p, so that it either equals p or is composite. Since we claimed P(x) is never composite, P(kp) = p for all k. That is, P(kp) - p is the zero polynomial, and hence P(x) must be constant.

seanieg89

Well-Known Member
I think I'm grasping at straws here, but:

P(kp) must be divisble by p, so that it either equals p or is composite. Since we claimed P(x) is never composite, P(kp) = p for all k. That is, P(kp) - p is the zero polynomial, and hence P(x) must be constant.
Bingo , back yourself more!

Will let someone else post one now.

-insert title here-
This isn't English.

Here, we say "Prove"

Anyway, I digress.

$\bg_white \noindent Do there exist infinitely many f : \mathbb{R} \mapsto \mathbb{R} such that \\f(x+y) = f(x)f(y)f(xy) \, ?$

seanieg89

Well-Known Member
This isn't English.

Here, we say "Prove"
Condescending much? Let people choose their own words. Clarity and precision of semantic content is far more important in mathematical writing than such choices in wording. Especially in a forum post of a single question lol.

I did not write this question as "Prove X" simply because I wanted students to approach the question without knowing for sure whether X was true or false.

-insert title here-
Condescending much? Let people choose their own words. Clarity and precision of semantic content is far more important in mathematical writing than such choices in wording. Especially in a forum post of a single question lol.

I did not write this question as "Prove X" simply because I wanted students to approach the question without knowing for sure whether X was true or false.
You could also say "prove your result" or "prove or disprove"

I see the second one a lot in Putnam questions.

seanieg89

Well-Known Member
You could also say "prove your result" or "prove or disprove"
I could. I could also say: "Justify your answer by way of proof", or countless variations of this.

I made a choice, and the meaning was unambiguous. You will be frequently frustrated in future studies if such variations in wording bother you.

InteGrand

Well-Known Member
Condescending much? Let people choose their own words. Clarity and precision of semantic content is far more important in mathematical writing than such choices in wording. Especially in a forum post of a single question lol.

I did not write this question as "Prove X" simply because I wanted students to approach the question without knowing for sure whether X was true or false.
I think Paradoxica might have been being tongue-in-cheek when he said that comment. Or at least, I thought he was being tongue-in-cheek when I first saw the comment (thought he was just jokingly taking a jab at English maybe), not sure now upon seeing his reply to your post I quoted.

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RealiseNothing

what is that?It is Cowpea
I think Paradoxica might have been being tongue-in-cheek when he said that comment. Or at least, I thought he was being tongue-in-cheek when I first saw the comment (thought he was just jokingly taking a jab at English maybe), not sure now upon seeing his reply to your post I quoted.
Nah he was defs srs lol.

seanieg89

Well-Known Member
Also, there are only finitely many solutions to that functional equation I believe.

It is clear that if a solution vanishes somewhere it vanishes everywhere (let y be a root to make the RHS vanish and then vary x so the LHS vanishing tells you that f is identically zero).

Let us assume now that f is a nonvanishing solution.

Applying the functional equation tells us:

f(x+y+z)=f(x+(y+z))=f(x)f(y+z)f(xy+xz)=f(x)f(y)f(z)f(yz)f(xy)f(xz)f(x^2yz).

On the other hand, if we group the terms differently, we obtain f(x+y+z)=f(y+(x+z))=f(x)f(y)f(z)f(yz)f(xy)f(xz)f(xy^2z).

Setting z=1 and dividing out the factors (which are nonzero!) tells us f(x^2y)=f(xy^2) for all real x and y.

But for nonzero a,b we can always solve x^2y=a, xy^2=b. (x^3=a^2/b, y^3=b^2/a).

This implies that f must be constant on R \ {0}. Moreover, by setting x = y =/= 0 in the functional equation, we obtain that this constant must satisfy x^3=x, which has only three solutions.

Similarly, setting x=y=0 tells us that the value at the origin must also solve this same cubic.

So we have narrowed down possible solutions to the functional equation to a finite set of functions and we are done.

(It would not take much additional effort to find ALL solutions to the functional equation if one so desired.)

seanieg89

Well-Known Member
Here's a cute one:

Find all solutions f: R\{2/3}->R to:

504x-f(x)=f(2x/(3x-2))/2.

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-insert title here-
Here's a cute one:

Find all solutions f: R->R to:

$\bg_white 504x-f(x)=\frac{f(\frac{2x}{3x-2})}{2}$
$\bg_white \noindent Observe g(x) = \frac{2x}{3x-2} is an involution. So if f(x) is a solution, then f(\frac{2x}{3x-2}) is a solution. Substituting in x \mapsto \frac{2x}{3x-2}, we have\\ \frac{1008x}{3x-2}-f(\frac{2x}{3x-2}) = \frac{f(x)}{2} \\ Solving simultaneously, we get 1008x(1 - \frac{1}{3x-2}) = \frac{3}{2}f(x) \\ 1008x (\frac{3(x-1)}{3x-2}) = \frac{3}{2}f(x) \\ f(x) = \frac{2016x (x-1)}{3x-2} \\ Substituting in x \mapsto \frac{2x}{3x-2} gives us f(x) = \frac{1008x (2-x)}{3x-2} \\ Substituting again returns us back to where we started. Cycling through the functional input doesn't give us anything different. This verifies all the possible solutions to the functional equation. By substitution, we see both functions satisfy the functional equation. \\ So the only two solutions are:\\f(x) = \frac{1008x (2-x)}{3x-2} \\ f(x) = \frac{2016x (x-1)}{3x-2}$

seanieg89

Well-Known Member
I do not believe your first solution is correct, neither is the statement that if f(x) is a solution then f(2x/(3x-2)) is one too.

This would be the case if the functional equation was symmetric in these two unknowns, but the factor of 1/2 stops that from happening.

A quick way to verify that this solution doesn't work without plugging into the functional equation directly is by considering the limit L of f(x)/x as x->0. The functional equation implies that L must be 1008 if it exists, whereas the limit in your first proposed solution is -1008.

The second solution is the unique solution and you are done after solving the simultaneous equations. (Because all of the previous lines are quantified over all x, you can actually be sure that this is the unique solution after reaching the line f(x)=blah).

(I obviously cooked up the numbers so that 2016 featured in the unique solution.)

-insert title here-
I do not believe your first solution is correct, neither is the statement that if f(x) is a solution then f(2x/(3x-2)) is one too.

This would be the case if the functional equation was symmetric in these two unknowns, but the factor of 1/2 stops that from happening.

A quick way to verify that this solution doesn't work without plugging into the functional equation directly is by considering the limit L of f(x)/x as x->0. The functional equation implies that L must be 1008 if it exists, whereas the limit in your first proposed solution is -1008.

The second solution is the unique solution and you are done after solving the simultaneous equations. (Because all of the previous lines are quantified over all x, you can actually be sure that this is the unique solution after reaching the line f(x)=blah).

(I obviously cooked up the numbers so that 2016 featured in the unique solution.)
Ah. Knew I should have just done a simple substitution instead of the involution property.

Was that intentional, by the way?

seanieg89

Well-Known Member
Ah. Knew I should have just done a simple substitution instead of the involution property.

Was that intentional, by the way?
Yep, you got the key idea, just a silly error.

Was what intentional?

-insert title here-
Yep, you got the key idea, just a silly error.

Was what intentional?
Involution

seanieg89

Well-Known Member
Involution
Oh yes, entirely. That was why I considered it cute, I quite like simple ideas exploiting symmetries like that.

seanieg89

Well-Known Member
Find with proof, all continuous functions f:R->R such that:

f(f(f(x)))=x for all real x.