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Question:

Aussme that the function $\color{blue}{f:N^{+}\to N^{+}}$,foy any $x\neq y$,if $\color{red}{p|f(x)+f(y)}$,then we have $\color{blue}{p|x+y}$, find $\color{red}{f(1)}$

where $p$ is prime number

Idea: Assmue $p=2$,if $f(x)+f(y)$ is even,then $x+y$ also even number.

Assmue that $p=3$, if $f(x)+f(y)=3k$,then $x+y=3m,k,m\in N^{+}$.

But How to determine the value $\color{red}{f(1)}$

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  • $\begingroup$ This holds for any prime number $p$? Please make the quantification explicit so one can tell which variables are picked dependent on which others. $\endgroup$ Jul 18, 2016 at 5:04

2 Answers 2

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Note that if $x+y=p^n$, then $p$ is the only allowed prime divisor of $f(x)+f(y)$. $f(x)+f(y)$ can't be $1$, so $p|f(x)+f(y)$.

Now consider the values of $f(1),f(2),...,f(p)$ mod $p$ (with $p>2$). We must have $p|f(1)+f(p-1)$, $p|f(2)+f(p-2)$, etc. If for some $i<j$ both less than or equal to $p$, $f(i)\equiv f(j)$ mod $p$, then $p|f(p-i)+f(j)$ but $p\nmid p-i+j$, a contradiction. Therefore $f(1),f(2),...,f(p)$ are all distinct mod $p$.

Now if $f(p^2-p)\equiv f(i)$ for some $i<p$, then $p|f(p^2-p)+f(p-i)$ but $p\nmid p^2-i$, a contradiction. Therefore $f(p^2-p)\equiv f(p)$ mod $p$.

But by the note above, $p|f(p^2-p)+f(p)$, so $p|2f(p)$. Therefore $p|f(p)$ and $p\nmid f(i)$ whenever $p\nmid i$. In particular, no odd prime divides $f(1)$, $f(2)$, or $f(4)$, so these three values must all be powers of $2$.

$f(1)+f(2)$ and $f(1)+f(4)$ are both odd,. If $f(1)$ is even, both $f(2)$ and $f(4)$ must be odd. Because they are powers of $2$, they would both have to be $1$, which is impossible. Therefore $f(1)$ is odd, so it must be $1$.

(At this point, we know that no prime can ever divide $f(x+1)-f(x)$, so the difference between two consecutive values of $f$ is $\pm1$. However, $f$ can never take the same value twice. The only possible $f$ is the identity function.)

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Here's a thought: the identity map, which sends every number to itself, satisfies this property. If you could prove that it is the only such function, you could get the result $f(1) = 1$.

From a meta perspective, we still have the following weaker result:

If this problem has a well-defined answer — meaning that we don't need to know anything more about the function $f$ because every function with these properties assigns the same value to $f(1)$— then the answer must be $f(1)=1$.

The reason is because we know that the identity map has these properties, and it assigns $f(1)=1$.

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  • $\begingroup$ yes,the same as mean,But How find the $f(1)$ $\endgroup$
    – math110
    Jul 18, 2016 at 1:57
  • $\begingroup$ I don't really see how this answers the question. $\endgroup$
    – user296602
    Jul 18, 2016 at 2:18

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