Proving a mathematical statement

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The discussion centers on proving the statement that there exists a positive integer m such that for all positive integers n, n+m is divisible by 3. A participant reflects on a similar problem where they successfully proved a related statement by letting m depend on n, but struggles to understand why reversing the condition makes the original statement false. It is clarified that in the original statement, m must be a fixed constant, while in the reversed statement, m can vary with n. This distinction in the dependency of m on n is crucial for understanding the validity of the statements. The conversation emphasizes the importance of the order of quantifiers in mathematical proofs.
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Homework Statement


There exists a number m, which is an element of the positive integers, that for all positive integers n, n+m can be divided by 3. Prove whether this statement is true or false.

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The Attempt at a Solution


I ran into a similar question earlier on, which just had the initial part reversed (as in, for all positive integers n there is a positive integer m so that n+m is divisible by 3). I proved that statement by letting m = 2n, and then 3n / 3 = n, which is a positive integer, proving that n+m is divisible. However, I don't understand why reversing the initial condition suddenly makes the entire statement false. Can I not do the same m = 2n idea to prove this statement? Could anyone explain why this is? Thank you in advance.
 
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m cannot depend on n here.
If such an m would exist, you would have to be able to say "m=1245" for example.
 
Ohh, so you're saying that m would be like a constant value, whereas in the other case it could be a variable? How exactly does the order play into this?
 
The order is in the statement.

"There is a number m [such] that for all integers n, ..." => fixed m, and then for all integers n something has to be true.
"For every n there is an integer m" => m can depend on n.
 
Ohh, I see, thank you so much for your help.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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