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Question:

For any $a,b\in \mathbb{N}^{+}$, if $a+b$ is a square number, then $f(a)+f(b)$ is also a square number. Find all such functions.

My try: It is clear that the function $$f(x)=x$$ satisfies the given conditions, since: $$f(a)+f(b)=a+b.$$

But is it the only function that fits our needs?

It's one of my friends that gave me this problem, maybe this is a Mathematical olympiad problem. Thank you for you help.

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    $\begingroup$ Well, for each $k \in \mathbb N^+$ there is the function $f(x) = k^2x$. $\endgroup$
    – TonyK
    Apr 28, 2014 at 11:35
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    $\begingroup$ It is proven that if $f$ is polynomial then $f(n)=a^2n$ or $f(n)=a^2/2$ for some constant $a$; see the article :On the sum-pth-power polynomial by Lwins G $\endgroup$
    – Elaqqad
    Feb 20, 2015 at 21:03
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    $\begingroup$ Let $A$ be a subset of $\mathbb{N}^+$ not containing two elements with perfect square sum. For instance, $A$ can be the set of natural numbers of the form $3n+1$. Define $f(n)$ to be $1$ if $n\in A$ and $8$ otherwise. $f$ is a solution. $\endgroup$ Jul 29, 2015 at 19:31
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    $\begingroup$ A generalization: Let $A$ be like I said. Define $f(n)$ to be $l^2-2k^2$ if $n\in A$ and $2k^2$ otherwise (where of course $2k^2<l^2$). $f$ is a solution. $\endgroup$ Jul 29, 2015 at 19:48
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    $\begingroup$ Perhaps I am being pessimistic, but to me there seem to be far too many solutions, none of which resemble each other in anyway, for this question to be manageable without further restriction - the condition on $f$ as it stands is very weak. To me it feels likely that your friend thought they solved it but missed something, or forgot additional info when they passed on the problem. $\endgroup$
    – ocg
    Dec 26, 2015 at 21:44

2 Answers 2

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It's not a complete answer, but as mentioned in comments, this problem probably missed some restrictions, and so have too many solutions. Thus I decided to answer this question for the case that $f$ have constant value in infinite (or finite by little changes) partition of $\mathbb N$.
I expect another answers for remained cases e.g when $f$ is an increasing function (polynomial case mentioned in comments).

Let $A$ is an infinite subset of $\mathbb N$, not containing two numbers with square sum (like https://oeis.org/A203988 except elements of the form $\frac{(2k)^2}{2}$ in this sequence) and $A'=\mathbb N -A$ . Suppose $A_1,A_2,...$ is an infinite non-empty partition of $A$, now $f$ could be defined as below
$$ f(n) = \begin{cases} a=\frac{(2k)^2}{2}& \quad \text{if } n \in A' \\a_1^2-a & \quad \text{if } n \in A_1\\a_2^2-a & \quad \text{if } n \in A_2\\.\\.\\. \end{cases} $$ where $k$ and $a_i \in \mathbb N$ .

Now if $x,y \in \mathbb N$ and $x+y$ is a perfect square, then both of $x$ and $y$ should be contained in $A'$, or on of them is in $A'$ and another one is in $A$ (and so contained in one of the $A_i$), in both cases $f(x)+f(y)$ is a perfect square .

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we can find some other solutions such the zero function or the functions under the form $f(n)=a^2n$ or the constant functions under the form $f(n)=\dfrac{a^2}{2}$, or an other function $f$ which can be determined following this method.
Here I am just giving a way to build a function with this property
suppose that $a, b \in \mathbb{N}^{+} $ such that their sum is a perfect square, that is there exists $n \in \mathbb{N}_{≥2}$ such that
$$ a+b=n^2 $$ which is equivalent to
$$ a=n^2-b $$ which implies that
$$ f(a)+f(b)=f(n^2-b)+f(b) $$ then from here we can see that we are looking for a function $f$ such that
$$ \forall n \in \mathbb{N}_{\geq 2}, \forall b \in { 1, 2,..., [ \dfrac{n^2}{2}] }, \exists a(b,n) \in \mathbb{N}; f(n^2-b)+f(b)=(a(b,n))^2$$ Let's start building
for $n=2$ we need to give values to $f(b) $ and $a(b, 2)$ for $ b \in \{ 1, 2 \}$ such that
$$ f(4-b)+f(b)=(a(b,2))^2$$ that is we need to give values to $f(1), f(2), f(3)$ and $a(1, 2), a(2,2)$ such that
$$ \begin{cases} f(3)+f(1)=(a(1,2))^2 \\ f(2)+f(2)=(a(2,2))^2 \end{cases} $$ Then we find that we can associate a random values to $f(1), a(1,2), a(2,2)$ and we define $f(2), f(3)$ as follows
$$ \begin{cases} f(2)=\dfrac{(a(2,2))^2}{2} \\ f(3)=(a(1,2))^2-f(1) \end{cases}$$ before moving to the next step we need to note that if $n$ is even and by taking $ b= \left[ \dfrac{n^2}{2}\right]$ the equality $ f(n^2-b)+f(b)=(a(b,n))^2 $ gives
$$ f( [ \dfrac{n^2}{2}] )=\dfrac{a( [ \dfrac{n^2}{2}], n)^2}{2}...(*) $$ and we can choose any natural value for $a( [ \dfrac{n^2}{2}], n)$ to define $f( [ \dfrac{n^2}{2}] )$.
in the next steps we have to take in account that $f( [ \dfrac{n^2}{2}])$ has been already determined.
for $n=3$ we need to give values to $f(b) $ and $a(b, 2)$ for $ b \in \{ 1, 2, 3, 4\}$ such that
$$ f(9-b)+f(b)=(a(b,3))^2$$ that is $$ \begin{cases} f(8)+f(1)=\dfrac{(a(1,3))^2}{2} \\ f(7)+f(2)=(a(2,3))^2\\ f(6)+f(3)=(a(3,3))^2\\ f(5)+f(4)=(a(4,3))^2 \end{cases}$$ Here we need to see that $f(2)$ and $f(8)$ have been already choosed from the equality $(*) $ that implies that $f(1)$ is not random as we find in the first step so we need to change the choice of $f(1)$ to be under the form $$ f(1)=\dfrac{(a(1,3))^2}{2}-f(8)$$ and since $f(2), f(3)$ are well determined we can easily associate values to $f(7), f(6)$.
but $f(4), f(5)$ remain not determined, it's clear that we can associate a random value to $f(4)$ and then we get $f(5)=(a(4,3))^2-f(4)$, but is this true?
Maybe in the next steps we will need to change one of these values as we have done with $f(1)$, and we iterate counting the next values of the function $f$.

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