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Extracurricular Elementary Mathematics Marathon (1 Viewer)

Paradoxica

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Good, you have done the harder part of the question :).

It remains to determine the exact solutions of g(x)=x but you are super close.
And you said it was harder. :p

I don't see how I can possibly bash this question.

Trying to come up with some way to restrict the solutions but nothing I think of will work.
 

seanieg89

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And you said it was harder. :p

I don't see how I can possibly bash this question.

Trying to come up with some way to restrict the solutions but nothing I think of will work.
I still think it is.

You will probably slap your head when you spot it :).
 

Paradoxica

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I still think it is.

You will probably slap your head when you spot it :).
Well I'm not sure if my mathematical rigor is exact, but here goes.

Suppose we had a solution with more than one of the numbers not equal to zero. However, if we multiply all the denominators up onto the other side, we would have an arbitrarily large set of congruences that are all equal to zero on the right hand side, but not equal to zero on the left hand side, as each term on the left hand side is not divisible by one of the primes on the right hand side. this leads to an impossible set of residues, and we have a contradiction. Thus, the only solutions that can exist are for exactly one of the integers equal to it's denominator, and every other term equal to zero. This verifies the class of solutions I mentioned earlier, and we are done.
 

seanieg89

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Exactly, although worded slightly funny. The point is that the j-th term on the LHS MUST be divisible by p_j as every other term trivially is.

This means that each a_j must be a non-negative integer multiple of p_j. (Because we do not get a prime factor of p_j in this term from multiplying out the denominators.)

I.e we have a finite ordered collection of non-negative integers summing to 1. The only way for this to happen is for one of these integers to be 1 and the others to be zero.

This lets us draw the desired conclusion.
 

KingOfActing

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I'm pretty new to this stuff, but I think I can do 3.



edit: Now that I think about it, Euler's Theorem is total overkill. Just note that x can be congruent to -2, -1, 0, 1 or 2 mod 5. After dealing with the 0 case, the +-1 cases have x^4 = 1, and the +-2 cases have x^4 = 16 which is congruent to 1 mod 5.

Also, since this is supposed to be a an elementary thread, if anyone hasn't done modular arithmetic yet, note that it is sufficient to make sure the statement holds true for x = 0,1,...,14. I don't think it can get any more elementary than that.
 
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seanieg89

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This establishes the claim for all n larger than 6.

Verifying the inequality for the five values of n between 2 and 6 with a calculator or by hand completes the proof.
 

Paradoxica

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This establishes the claim for all n larger than 6.

Verifying the inequality for the five values of n between 2 and 6 with a calculator or by hand completes the proof.
Wow... you're so bothered to type up all that ... My solution consisted of creating an intermediate inequality and inducting on that.
 

seanieg89

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Wow... you're so bothered to type up all that ... My solution consisted of creating an intermediate inequality and inducting on that.
Just went with the immediate idea that popped into my head, didn't take long to tex/compute.

If you want a weaker constant than 1/sqrt(3) (like 1/sqrt(2) for instance) then the same method as above gets you immediate results without needing to go several terms into the product before doing the telescoping.
 

Paradoxica

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Just went with the immediate idea that popped into my head, didn't take long to tex/compute.

If you want a weaker constant than 1/sqrt(3) (like 1/sqrt(2) for instance) then the same method as above gets you immediate results without needing to go several terms into the product before doing the telescoping.
The intermediate inequality you induct on is placing 1/sqrt(3n+1) between the two. It may seem arbitrary but the problem is that the right hand side doesn't shrink fast enough for an inductive proof to be able to compare the two sides.
 

seanieg89

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You should probably specify the domain of the functional equation you posted Paradoxica. It it the reals?


Assuming this is the case (and in fact this working is also valid if the domain is the integers, rationals or complex numbers), then:

6/

Let x=0 and let y=-f(0)+z, so

f(z)=f(f(0)+y)=y=-f(0)+z.

This implies that all solutions are of the form f(z)=z+c for some constant c.

However, if we apply this formula to our original functional equation, we get:

x+y+2c=x+y.

Hence c must be zero, and the only solution to the functional equation is the identity function f(x)=x.
 
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seanieg89

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Lols, mega procrastination on my behalf. Am supposed to be brushing up on something for a supervisor meeting. These questions are fun though.
 

seanieg89

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

Justify your response.
 
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