An FE. Who woulda thunk it?

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1943 posts

#1 h Apr 15, 2021, 5:13 PM • 9 Y

Y by samrocksnature, tastymath75025, FaThEr-SqUiRrEl, icematrix2, megarnie, mathlearner2357, jhu08, Lamboreghini, ImSh95

Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ $f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Orestis_Lignos

555 posts

Orestis_Lignos

#2 h Apr 15, 2021, 5:13 PM • 8 Y

Y by samrocksnature, FaThEr-SqUiRrEl, icematrix2, megarnie, jhu08, ImSh95, ike.chen, Amir Hossein

Claim: $f^2(m^2+n^2)=f(2mn)f(m^2-n^2)$ for all $m>n$ .
Proof: The RHS equals $f((m^2-n^2)^2+(2mn)^2) = f((m^2+n^2)^2)=f^2(m^2+n^2)$ hence done $\blacksquare$ .

Therefore $f(m)=f(n)=1$ implies $f(2mn)=f(m^2-n^2)=1$ as well.

Note that $f(1)=1$ by the first condition and $f(2)=1$ by the second.

Now assume that for all integers $m$ up to $k \geq 2$ , $f(m)=1$ holds.

If $k+1 \equiv 2 \pmod 4$ , take $m,n$ such that $2mn=k+1$ and note that $m,n \leq \frac{k+1}{2} <k$ so $f(m)=f(n)=1$ by the inductive hypothesis hence $f(2mn)=f(k+1)=1$ , so we are done.

If it is not $2 \pmod 4$ we may take $m>n$ such that $m^2-n^2=k+1$ and note that $2n<2m=(m-n)+(m+n) \leq (m-n)(m+n)+1=k+2 \leq 2k,$ hence both $m,n$ are $\leq k$ so $f(m)=f(n)=1$ by the inductive hypothesis, therefore again $f(m^2-n^2)=f(k+1)=1$ .

Thus, $f \equiv 1$ for all positive integers, which evidently satisfies, so it is our only solution.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

smartkmd1

132 posts

smartkmd1

#3 h Apr 15, 2021, 5:16 PM • 14 Y

Y by samrocksnature, FaThEr-SqUiRrEl, yk2007, icematrix2, lovefractions, math_genius_11, SSaad, mathking999, Moha_b, ImSh95, EpicBird08, Mango247, Mango247, Mango247

We claim the only solution to the equation is $\boxed{f(x) = 1}$ .

This works because:
$f(a^2+b^2) = f(a)f(b) \Rightarrow 1=1\cdot1$ $f(a^2) = f(a)^2 \Rightarrow 1 = 1^2$ We will prove this using strong induction:

Base Case: $n=1,2$
$f(1^2) = f(1)^2$ $f(1)\left(f(1) - 1\right) = 0$ However, since $0$ is not in $\mathbb N$ , $f(1) = 1$ .
Now, plugging in $a=b=1$ , into the first equation:
$f(2) = f(1)f(1) = 1$
Inductive Step:
Suppose $f(x) = 1$ for all integers $0<x<n$ .
Then, we will prove that $f(n) =1$ .
First, plug in $a=2uv$ and $b=u^2-v^2$ into the first equation:
$f(2uv)f(u^2-v^2) = f\left((2uv)^2 + (u^2-v^2)^2\right) = f((u^2+v^2)^2) = f(u^2+v^2)^2$ However, plugging in $a=u$ and $b=v$ :
$f(u^2+v^2) = f(u)f(v)$ $f(u)^2f(v)^2=f(u^2+v^2)^2$ This means that:
$f(2uv)f(u^2-v^2) = f(u)^2f(v)^2 \tag{$\ast$}$
Now, we take casework on the parity of $n$ :

Case 1: $n$ is even
Let $n=2k$ . Then, plugging in $u=k$ and $v=1$ into $(\ast)$ ,
$f(2k)f(k^2-1)=f(k)^2f(1)^2$ Since $k<n$ , by inductive hypothesis:
$f(n)f(k^2-1)=1$ However, since the range of $f$ is $\mathbb N$ , both the terms on the left hand side have to be $1$ . This means that $f(n)=1$ , so we are done with this case.

Case 2: $n$ is odd
Let $n=2k+1$ . Then, plugging in $u=k+1$ and $v=k$ into $(\ast)$ ,
$f(2k(k+1))f((k+1)^2-k^2)=f(k)^2f(k+1)^2$ $f(2k(k+1))f(n)=f(k)^2f(k+1)^2$ Again, since $k,k+1<n$ , by inductive hypothesis:
$f(2k(k+1))f(n) = 1$ Again, since the range of $f$ is $\mathbb N$ , both the terms on the left hand side have to be $1$ which means that $f(n)=1$ , so we are done with this case as well.
Therefore, both cases yield $f(n)=1$ , so we are done with our inductive step.

Our induction covers the entire domain of $f$ , so we can conclude that $f(x)=1$ is the only solution.

This post has been edited 1 time. Last edited by smartkmd1, Apr 15, 2021, 5:18 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

CT17

1481 posts

CT17

#5 h Apr 15, 2021, 5:27 PM • 12 Y

Y by samrocksnature, FaThEr-SqUiRrEl, v4913, tenebrine, icematrix2, SSaad, centslordm, megarnie, mathking999, Bakhtier, ImSh95, ike.chen

Solution with some details left out

This post has been edited 2 times. Last edited by CT17, Apr 15, 2021, 5:39 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

rocketracoon

272 posts

rocketracoon

#6 h Apr 15, 2021, 5:31 PM • 6 Y

Y by samrocksnature, FaThEr-SqUiRrEl, icematrix2, Cusofay, megarnie, ImSh95

Was this possible without splitting into even/odd? I'm did something weird and now I don't know if it's legit.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

amuthup

779 posts

amuthup

#7 h Apr 15, 2021, 5:36 PM • 6 Y

Y by samrocksnature, FaThEr-SqUiRrEl, aie8920, icematrix2, megarnie, ImSh95

Feel like the induction idea is pretty standard.

The only solution is $\boxed{f\equiv 1},$ which clearly works. To prove that $f(n)=1$ for all $n,$ we will induct on $n.$

$\emph{Base Case: }$ $n=1$

Plugging $a=1$ into the second assertion yields $f(1)=f(1)^2,$ so since $f(1)$ is a natural number, it must be $1.$

$\emph{Inductive Step: }$

Case 1: $n$ is odd

Plugging $(a,b)=(n,\frac{n^2-1}{2})$ into the first assertion gives $f\left(\frac{n^4+2n^2+1}{4}\right)=f(n)f\left(\frac{n^2-1}{2}\right).$ Since $\frac{n^4+2n^2+1}{4}=(\frac{n^2+1}{2})^2,$ the second assertion yields $f\left(\frac{n^2+1}{2}\right)^2=f(n)f\left(\frac{n^2-1}{2}\right).$ On the other hand, by plugging $(a,b)=(\frac{n-1}{2},\frac{n+1}{2})$ into the first assertion, we obtain $f\left(\frac{n^2+1}{2}\right)=f\left(\frac{n-1}{2}\right)f\left(\frac{n+1}{2}\right),$ which equals $1$ by the inductive hypothesis.

Thus, $f(n)f\left(\frac{n^2-1}{2}\right)=1,$ forcing $f(n)=1.$
Case 2: $n$ is even and $n\ge 4$

Plugging $(a,b)=(n,\frac{n^2-4}{4})$ into the first assertion gives $f\left(\frac{n^4+8n^2+16}{4}\right)=f(n)f\left(\frac{n^2-4}{4}\right).$ Since $\frac{n^4+8n^2+16}{16}=(\frac{n^2+4}{4})^2,$ the second assertion yields $f\left(\frac{n^2+4}{4}\right)^2=f(n)f\left(\frac{n^2-4}{4}\right).$ On the other hand, plugging $(a,b)=(\frac{n}{2},1)$ into the first assertion, we get $f\left(\frac{n^2+4}{4}\right)=f\left(\frac{n}{2}\right)f(1),$ which equals $1$ by the inductive hypothesis.

Hence $f(n)f(\frac{n^2-4}{4})=1,$ so $f(n)$ must be $1.$
Case 3: $n=2$

Plugging $(a,b)=(1,1)$ into the first assertion gives $f(2)=f(1)^2=1,$ as desired.

Our proof is complete.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

vvluo

1574 posts

vvluo

#8 h Apr 15, 2021, 5:58 PM • 7 Y

Y by mathleticguyyy, samrocksnature, FaThEr-SqUiRrEl, icematrix2, megarnie, ImSh95, Mango247

Sketch: f \cong 1

get f(1)=f(2)=1

If f(n) =f(n-1)=1, then we have f(2n)=f(2n-1)=1 for all n ge 2

Hinges on (n^2 + 1)^2 = (n^2 -1)^2 + (2n)^2

and (2n^2 -2n+1)^2=(2n-1)^2 + (2n^2 - 2n)^2

Way too hard for a J1 imo

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

IAmTheHazard

5000 posts

IAmTheHazard

#9 h Apr 15, 2021, 6:05 PM • 7 Y

Y by samrocksnature, FaThEr-SqUiRrEl, icematrix2, megarnie, ImSh95, mathmax12, Mango247

This was hard especially for a j1.
There are basically three steps:
Step 1: prove $f(1)=f(2)=1$ (easy)
Step 2: Prove that $f(a)=1 \implies f(2a)=1$
Step 3: Prove that $f(a)=f(a+1)=1 \implies f(2a+1)=1$ .
Then strong induction finishes. However neither step 2 nor step 3 are particularly easy to find, and it's hard to naturally "get" there without taking a fairly big leap of faith.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

lethan3

907 posts

lethan3

#10 h Apr 15, 2021, 6:09 PM • 10 Y

Y by samrocksnature, FaThEr-SqUiRrEl, icematrix2, mathleticguyyy, megarnie, ImSh95, Mango247, Mango247, Mango247, dbnl

vvluo wrote:

Way too hard for a J1 imo

you can say that again i wasted the whole 4.5 hours on this

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Awesome_guy

862 posts

Awesome_guy

#11 h Apr 15, 2021, 6:23 PM • 8 Y

Y by samrocksnature, FaThEr-SqUiRrEl, icematrix2, ImSh95, Mango247, Mango247, Mango247, a_smart_alecks

Very contrived :noo:

.
$f(1)=1$ , $f(2)=1$ . It can be confirmed that $f(k)=1$ for all $1\leq k\leq 13$ . It suffices to show that for all $1\leq i\leq j$ such that $f(i)=1\implies f(j+1)=1$ . Note that $(m+n)^2+(8m+3n)^2=(7m+3n)^2+(4m+n)^2$ , and by Frobenius coin all integers greater than $13$ can be expressed as $8m+3n$ , as desired.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

rocketracoon

272 posts

rocketracoon

#12 h Apr 15, 2021, 6:26 PM • 5 Y

Y by samrocksnature, FaThEr-SqUiRrEl, icematrix2, megarnie, ImSh95

My solution was so weird...
Proved that if f(x+1)=1, then f(x)=1.
Then infinite values give f(y)=1, so we're done

This post has been edited 1 time. Last edited by rocketracoon, Apr 15, 2021, 6:27 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

TwilightZone

128 posts

TwilightZone

#14 h Apr 15, 2021, 7:19 PM • 5 Y

Y by FaThEr-SqUiRrEl, samrocksnature, icematrix2, megarnie, ImSh95

bump plsss sry idk my solution looks different so im thinking there is some error somewhere and idk where

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

brainiacmaniac31

2170 posts

brainiacmaniac31

#15 h Apr 15, 2021, 7:22 PM • 4 Y

Y by FaThEr-SqUiRrEl, samrocksnature, icematrix2, ImSh95

I claim the only answer is $\boxed{f(n)=1}$ for all $n\in\mathbb{N}$ . It is easy to verify that this satisfies the problem conditions.

First, note that $f(1)=1$ because $f(1)=f(1)^2$ . Furthermore, $f(2)=1$ because $f(2)=f(1^2+1^2)=f(1)^2=1$ .

Observe that
$f\big((x^2-y^2)^2+(2xy)^2\big)=f\big((x^2+y^2)^2\big)=f(x^2+y^2)^2=f(x)^2f(y)^2$ $f\big((x^2-y^2)^2+(2xy)^2\big)=f(x^2-y^2)f(2xy)$ $\implies f(x)^2f(y)^2=f(x^2-y^2)f(2xy)\qquad (\clubsuit).$ In particular, if $f(x)=f(y)=1$ , then by $(\clubsuit)$ we know that $f(x^2-y^2)=f(2xy)=1$ . We now finish with strong induction.

We have already shown our base cases: $f(1)=f(2)=1$ . Now, assume that $f(k)=1$ for all $k\in[1,n]$ . If $n+1$ is even, then by our inductive assumption, $f(1)=f\left(\frac{n+1}{2}\right)=1$ , so by $(\clubsuit)$ , $f\left(2\cdot1\cdot\frac{n+1}{2}\right)=f(n+1)=1$ . If $n+1$ is odd, let $n=2k$ . By our inductive assumption, $f(k)=f(k+1)=1$ , so again by $(\clubsuit)$ , $f\big((k+1)^2-k^2\big)=f(2k+1)=f(n+1)=1$ . This completes the induction, so $f(n)$ is always equal to $1$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

i3435

1349 posts

i3435

#16 h Apr 15, 2021, 7:25 PM • 17 Y

Y by tastymath75025, Inconsistent, FaThEr-SqUiRrEl, Aryan-23, samrocksnature, icematrix2, Geometry285, IAmTheHazard, megarnie, centslordm, mathleticguyyy, pog, rayfish, ImSh95, fuzimiao2013, Quidditch, Sedro

The $f(a^2)=a^2$ condition is not necessary.

I will show $f(x)=1$ for every $x$ , this works.

Let $n\neq 1,2,3,4,5,6,8,10$ . Then $n>\frac{n+3}{2},\frac{n-5}{2}>0$ if $n$ is odd and $n>\frac{n+6}{2},\frac{n-10}{2}>0$ if $n$ is even.

For odd $n$ we have $n^2-(n-2)^2=(\frac{n+3}{2})^2-(\frac{n-5}{2})^2=4n-4$ , so $f(n)f(\frac{n-5}{2})=f(n-2)f(\frac{n+3}{2})$ .

For even $n$ we have $n^2-(n-4)^2=(\frac{n+6}{2})^2-(\frac{n-10}{2})^2=8n-16$ , so $f(n)f(\frac{n-10}{2})=f(n-4)f(\frac{n+6}{2})$ .

Thus for these $n$ strong induction works, we just have to bash out $n=1,2,3,4,5,6,8,10$ .

$1025=32^2+1^2=20^2+25^2,32=4^2+4^2,25=4^2+3^2,20=4^2+2^2$ . This means $f(1025)=f(4)^2f(1)=f(4)^2f(2)f(3)$ . However $f(2)=f(1)^2$ , so $f(1)=f(2)=f(3)=1$ .

$f(5)=f(2)f(1),f(8)=f(2)^2=1,f(10)=f(3)f(1)=1$ .

$325=18^2+1^2=17^2+6^2,18=3^2+3^2,17=4^2+1^2$ . This means that $f(325)=f(4)f(6)f(1)=f(3)^2f(1)=1$ , so $f(4)=f(6)=1$ , as desired.

Edit: $n=8$ works from the identity also, just use $\frac{10-n}{2}$ , but I was too lazy to point this out.

This post has been edited 2 times. Last edited by i3435, Apr 16, 2021, 7:33 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

tastymath75025

3223 posts

tastymath75025

#17 h Apr 15, 2021, 7:28 PM • 48 Y

Y by ghu2024, khina, SnowPanda, franchester, P_Groudon, tapir1729, vvluo, Aryan-23, insertionsort, i3435, FaThEr-SqUiRrEl, Lcz, v4913, spartacle, samrocksnature, VipMath, jeff10, Hamroldt, Rg230403, OlympusHero, tigerzhang, sunfishho, icematrix2, centslordm, ike.chen, bobthegod78, Geometry285, mathlearner2357, jhu08, RubixMaster21, megarnie, suvamkonar, asdf334, lethan3, pog, ImSh95, Flying-Man, rayfish, ConfidentKoala4, Quidditch, IMUKAT, BDRSDG, OronSH, mathmax12, the_mathmagician, Jack_w, Sedro, Turtwig113

My problem

I wrote it a few years ago by trying to integrate the characterization of Pythagorean triples into an FE, because if $a^2+b^2=c^2$ then you get $f(a)f(b)=f(c)^2$ . Picking the right triples $(a,b,c)$ then lets you induct to show $f\equiv 1$ , as in many of the above solutions.

Honestly a bit surprised it ended up as JMO #1 though, especially because nobody bothered to tell me the problem was accepted :gathering:

This post has been edited 1 time. Last edited by tastymath75025, Apr 15, 2021, 7:29 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mathmax12

5980 posts

mathmax12

#110 h Aug 20, 2023, 1:40 AM

Y by

We claim that $\boxed{f(x)=1}$ , is the only function that works.
Claim: $f^2(m^2+n^2)=f(2mn)f(m^2-n^2)$
Proof: It is obvious that $f((2mn)^2+(m^2-n^2)^2)=f((m^2+n^2)^2)=f^2(m^2+n^2)$ , hence, proved.
So $f(2mn)=f(m^2-n^2)=1.$
Now we use induction. The base case, always works.

oops i cant continue, i cant do the rest

This post has been edited 5 times. Last edited by mathmax12, Aug 20, 2023, 1:46 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Shreyasharma

666 posts

Shreyasharma

#112 h Sep 26, 2023, 4:50 PM

Y by

We claim the only satisfactory functions are $f(x) \equiv 1$ . It is easy to see this works.

Let $P(a, b)$ denote the assertion for the equation $f(a^2 + b^2) = f(a)f(b)$ and $Q(a)$ denote the assertion for the second equation.

Begin by noting from $Q(1)$ we have $f(1) = f(1)^2$ , giving $f(1) = 1$ . Now from a simple induction $f(2^k) = 1$ for all $k \geq 0$ . Indeed, for odd $k$ take $P\left(2^{\frac{k-1}{2}}, 2^{\frac{k-1}{2}}\right)$ , and for even $k$ take $Q\left(2^{\frac{k}{2}}\right)$ .

Now we prove the claim using strong induction. From $P(k^2+k, 2k + 1)$ gives,
$\begin{align*} f\left(\left((k+1)^2 + k^2\right)^2\right) &= f((k(k+1)) \cdot f(2k + 1)\\ f\left( (k+1)^2 + k^2 \right) ^2 &= f(k^2 + k) \cdot f(2k+1)\\ \left( f(k+1)f(k) \right)^2 &= f(k^2+k) \cdot f(2k+1)\\ 1 &= f(k^2+k) \cdot f(2k+1) \end{align*}$ Similarly from $P(2k + 2, k^2 + 2k)$ we have,
$\begin{align*} \left(f(k+1) \right)^2 &= f(2k+2) \cdot f(k^2 + 2k)\\ 1 &= f(2k+2) \cdot f(k^2 + 2k) \end{align*}$ It is now easy to see that we can show $f \equiv 1$ by solving for odd and even $x$ using the first and second equations respectively.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

joshualiu315

2513 posts

joshualiu315

#113 h Oct 17, 2023, 6:03 AM • 1 Y

Y by ATGY

We claim the answer is $\boxed{f(x)=1}$ . We will show this using a claim:

Claim: If $f(a)=f(b)=1$ , $a>b$ , we must have $f(a^2-b^2)=f(2ab)=1$ as well.

Proof: We have

$\begin{align*} 1=f(a)f(b)&=f(a^2+b^2)=f((a^2+b^2)^2) \\ &= f((a^2-b^2)^2+(2ab)^2)=f(a^2-b^2)f(2ab) \end{align*}$
so we must have $f(a^2-b^2)=f(2ab)=1$ . $\square$

We have $f(1)=f(2)=1$ , so a simple induction finishes.

If $n$ is even, write as $n=2k$ , and plug in $(a,b)=(k,1)$ , so that $2ab=n$ .
If $n$ is odd, write as $n=2k-1$ , and plug in $(a,b)=(k, k-1)$ so that $a^2-b^2=n$ .

This post has been edited 1 time. Last edited by joshualiu315, Oct 17, 2023, 6:10 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

shendrew7

787 posts

shendrew7

#114 h Dec 17, 2023, 8:56 PM • 1 Y

Y by GeoKing

It's easy to get $f(1)=f(2)=1$ . We can use our two conditions to get
$f(a) f(b) = \sqrt{f\left((a^2+b^2)^2\right)} = \sqrt{f\left((a^2-b^2)^2+(2ab)^2\right)} = \sqrt{f\left(a^2-b^2\right) f(2ab)}.$
Note that our function is over the natural numbers, so if $f(a)=f(b)=1$ , then we immediately know $f\left(a^2-b^2\right)=f(2ab)=1$ . We now use strong induction with our base cases 1 and 2 to show $f(x)=1$ for all natural $x$ . Assume $f(1)=f(2)= \ldots = f(k-1)=1$ for an integer $k>2$ , and we want to show $f(k)=1$ .

If $k$ is even: Substitute $\left(\frac k2, 1\right)$ .
If $k$ is odd: Substitute $\left(\frac{k+1}{2}, \frac{k-1}{2}\right)$ .

This alogirthm tells us $f(k)=1$ for all $k$ , so our solution is $\boxed{f(x)=1}$ , which clearly works.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

kamatadu

466 posts

kamatadu

#115 h Dec 31, 2023, 8:09 AM • 1 Y

Y by GeoKing

$f \equiv 1$ is the only solution.

Note that $f(a^2) = f(a)^2$ . Now by putting $a=1$ , we get $f(1) = 1$ .
$f(2) = f(1^2 + 1^2) = f(1) \cdot f(1) = 1$ .
$f(4) = f(2^2) = f(2)^2 = 1$ .
$f(5) = f(1^2 + 2^2) = f(1) \cdot f(2) = 1$ .

Now note that $f(3^2 + 4^2) = f(5^2) \implies f(3)f(4) = f(5)^2 \implies f(3) = 1$ .

$f(8) = f(2^2+2^2) = f(2) \cdot f(2) = 1$ .
$f(10) = f(1^2 + 3^2) = f(1) \cdot f(3) = 1$ .
$f(6^2 + 8^2) = f(10^2) \implies f(6) = 1$ .
$f(7^2 + 24^2) = f(25^2) = f(5)^4 = 1\implies f(7) = 1$ .

Now we proceed by using strong induction. We show that $f(4k+1) = f(4k+2) = f(4k+3) = f(4k+4) = 1$ . This would simply result in $f\equiv 1$ .

We have already prove our base cases for $k=0$ and $1$ .

Firstly, let $P(a,b)$ be the assertion for $f(a^2+b^2) = f(a)\cdot f(b)$ . So for sone $m>n$ , we have,
$\begin{align*} P(m^2-n^2,2mn) &\implies f((m^2-n^2)^2 + (2mn)^2) = f(m^2-n^2)\cdot f(2mn)\\ &\implies f((m^2+n^2)^2) = f(m^2-n^2)\cdot f(2mn)\\ &\implies f(m^2 + n^2)^2 = f(m^2 - n^2) \cdot f(2mn)\\ &\implies f(m)^2\cdot f(n)^2 = f(m^2 - n^2) \cdot f(2mn) .\end{align*}$
Denote the final equation above by the assertion $Q(m,n)$ .

$Q(2k+1,2k) \implies f(2k+1)^2 \cdot f(2k)^2 = f(4k+1)\cdot f(2(2k+1)2k) \implies f(4k+1) = 1$ due to divisibility.

$Q(2k+1,1) \implies f(2(2k+1)\cdot 1)\cdot f(2k(2k+2)) = 1 \implies f(4k+2) = 1$ .

$Q(2k+2,2k+1) \implies f(4k+3) = 1$ .

$Q(k+1,2) \implies f(2(k+1) \cdot 2) = 1\implies f(4k+4) = 1$ .

This completes our induction and we are done. :yoda:

This post has been edited 1 time. Last edited by kamatadu, Dec 31, 2023, 8:10 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

GrantStar

812 posts

GrantStar

#116 h Jan 24, 2024, 3:24 AM

Y by

The answer is $f\equiv 1$ . Check that this works, and it is easy to show for small $n$ .
Then, if $n=2k$ is even, $f(k^2-1)f(2k)=f\left((k^2+1)^2\right)=f(k)^2f(1)^2$ and if $n=2k+1$ is odd, $f(2k+1)f(2k^2+2k)=f(2k^2+2k+1)^2=f(k)^2f(k+1)^2$ and we are done by induction.

This post has been edited 3 times. Last edited by GrantStar, Jan 24, 2024, 3:25 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

AngeloChu

471 posts

AngeloChu

#119 h Mar 13, 2024, 3:35 PM

Y by

first, we can very easily prove $f(1) = 1$ , $f(a^2 + 1) = f(a)$ , and $f(2a) = f(a)$
we now proceed by recursion
if $f(p)=1$ , $f(p^2+1)=1$ , and $f(2p^2+2)=1$ , but $f(2p^2+2)=f(p+1)f(p-1)$ , so $f(p+1)=f(p-1)=1$
therefore, we get that all $f(x)=1$ , and that is our only solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

eibc

595 posts

eibc

#120 h Mar 13, 2024, 8:58 PM

Y by

We claim that the solution is $f \equiv 1$ , which works. We will prove that $f(1) = f(2) = \cdots = f(2n) = 1$ for any positive integer $n$ by strong induction on $n$ .

For our base case of $n = 1$ , we have $f(1)^2 = f(1)$ , so $f(1) = 1$ . We also find that $f(2) = f(1^2 + 1^2) = f(1)f(1) = 1$ as well.

For the inductive step, suppose that when $n=k$ , all of $f(1), f(2), \ldots, f(2k)$ equal $1$ ; we then wish to prove that $f(2k + 1) = f(2k + 2) = 1$ . We will need the following result:

Claim: For $a, b \in \mathbb{N}$ , if $f(a) = f(b) = 1$ and $a > b$ , then $f(2ab) = 2(a^2 - b^2) = 1$ .

Proof: Observe that $(a^2 - b^2)^2 + (2ab)^2 = (a^2 + b^2)^2$ . Therefore
$f(a^2 - b^2)f(2ab) = f(a^2 - b^2)^2 + (2ab)^2) = f((a^2 + b^2)^2) = f(a^2 + b^2)^2 = (f(a)f(b))^2 = 1,$ which means that $f(a^2 - b^2) = f(2ab) = 1$ .

Taking $a = k + 1$ and $b = k$ , we find that $f(2k + 1) = 1$ as $(k + 1)^2 - k^2 = 2k + 1$ . Taking $a = k + 1$ and $b = 1$ , we get $f(2k + 2) = 1$ . So, we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Mr.Sharkman

486 posts

Mr.Sharkman

#121 h Apr 1, 2024, 9:16 PM

Y by

Solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

pie854

243 posts

pie854

#122 h Apr 24, 2024, 7:49 PM

Y by

First of all, clearly $f(1)=1$ and $f(2)=1$ . Since $f(a^2)=f(2a^2)$ , we get $f(2a)=f(4a^2)=f(a^2)=f(a)$ [Edit: oops, this part is not correct...]. Also $f(2a+1)=f((2a+1)^2+1)=f(4a^2+4a+2)=f(2a^2+2a+1)=f(a^2+(a+1)^2)=f(a)f(a+1).$ So by induction it follows that $f(x)\equiv 1$ , which is indeed a solution.

This post has been edited 1 time. Last edited by pie854, Apr 24, 2024, 9:05 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Martin2001

126 posts

Martin2001

#123 h Apr 25, 2024, 9:44 PM

Y by

We show that $f \equiv 1$ by induction. Note that $f(2)=f(1)f(1)=1$ because $0$ is not in the domain. Also note that $f(a)f(b)=1$ means that $f(a)=f(b)=1$ as there are no other numbers in the domain that have a product of $1.$ Thus, splitting cases into evens and odds we have that
$\begin{align*} f(2K)f(k^2-1)=f(k^2+1)&=f(k)^2f(1)^2=1 \\ f(2k+1)f(2k^2+2k)=f(2k^2+2k+1)^2=f(k)^2f(k+1)^2=1. \end{align*}$ This completes the induction. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ihatemath123

3426 posts

ihatemath123

#124 h Aug 15, 2024, 4:32 PM

Y by

We claim the answer is $f(x) = 1$ only.

We will that $f(n) = 1$ with strong induction on $n$ :

Base case: Our base cases are $n=1$ and $n=2$ . The former follows from plugging $a=1$ into the second given equation, while the latter follows from plugging in $(a,b) = (1,1)$ into the first given equation.
Inductive step: For $n$ of the form $2k$ where $k > 1$ , we use the identity
$(2k)^2 + \left( k^2 -1 \right)^2 = \left( k^2 + 1 \right)^2$ which lets us go from $f(1) = 1$ and $f(k) = 1$ to $f(2k) = 1$ . For $n$ of the form $2k+1$ , we use the identity
$(2k+1)^2 + (2k^2 + 2k)^2 = \left( k^2 + (k+1)^2 \right)^2,$ which lets us go from $f(k) = 1$ and $f(k+1) = 1$ to $f(2k+1) = 1$ .

This post has been edited 2 times. Last edited by ihatemath123, Aug 15, 2024, 4:34 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Jndd

1417 posts

Jndd

#125 h Aug 22, 2024, 4:53 AM

Y by

We claim that the answer is $f(x)=1$ for all $x\in\mathbb{N}$ , and this solution clearly satisfies the equation. First, we have $f(1^2)=f(1)^2$ , giving $f(1)=1$ . We also have $f(2)=f(1^2+1^2)=f(1)^2=1$ Suppose that $a^2+b^2=c^2$ for some $a,b,c\in\mathbb{N}$ . Then, we get $f(c)^2=f(c^2)=f(a^2+b^2)=f(a)f(b).$ Using strong induction, we will prove $f(2n+1)$ and $f(2n+2)$ are both $1$ assuming $f(x)=1$ for $x\in\{1,2,\ldots, 2n\}$ . We have $1=(f(n)f(n+1))^2=f((n^2+(n+1)^2)^2)=f((2n+1)^2+(2n^2+2n)^2)=f(2n+1)f(2n^2+2n),$ giving $f(2n+1)=1$ . Similarly, we have $1=(f(n+1)f(1))^2=f(((n+1)^2+1^2)^2)=f((2n+2)^2+f(n^2+2n)^2) = f(2n+2)f(n^2+2n),$ giving $f(2n+2)$ , and this completes our induction, as desired.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

RetroFuel

11 posts

RetroFuel

#126 h Sep 9, 2024, 10:57 AM

Y by

We will prove that $\boxed{f\equiv1}$ .
We first use $P(1,1)$ to obtain $f(1)=1$ . We then use $P(a,1)$ to obtain that $f(a^2+1)=f(a)$ . Now we will show via strong induction that if $f(x)=1$ for $1\leq x \leq n$ then $f(n+1)=1$ . Base cases $n=1,2$ can be easily verified.

Case 1: If $n+1$ is odd
We will write $n+1$ as $2k+1$ . We can construct a Pythagorean triplet $\{ a,b,c\}=\{2k+1,2k^2+2k,2k^2+2k+1\}$ . From $f(2k+1,2k^2+2k)$ we get $f((2k+1)^2+(2k^2+2k)^2)=f(2k+1)f(2k^2+2k)$ $f((2k^2+2k+1)^2)=f(2k+1)f(2k^2+2k)$ $f(2k^2 + 2k + 1)^2=f(2k+1)f(2k^2+2k)$ $f(k^2 + (k+1)^2)^2=f(2k+1)f(2k^2+2k)$ $f(k)f(k+1)f(k)f(k+1)=f(2k+1)f(2k^2+2k)$ $1=f(2k+1)f(2k^2+2k)$ $\Rightarrow f(2k+1)=1 \Rightarrow \boxed{f(n+1) = 1}$ Case 2: If $n+1$ is even
We will write $n+1$ as $2k$ . We can construct a Pythaogrean triplet $\{a,b,c\}=\{k^2-1,2k,k^2+1\}$ . From $f(k^2-1,2k)$ we get
$f((k^2-1)^2+(2k)^2)=f(k^2-1)f(2k)$ $f(k^4+2k^2+1)=f(k^2-1)f(2k)$ $f((k^2+1)^2)=f(k^2-1)f(2k)$ $f(k^2+1)^2=f(k^2-1)f(2k)$ $f(k)^2=f(k^2-1)f(2k)$ $1=f(k^2-1)f(2k)$ $\Rightarrow f(2k)=1 \Rightarrow\boxed{f(n+1)=1}$