Let N denote the set of positive integers. Find all functions f:N→N such that for positive integers a and b, f(a2+b2)=f(a)f(b) and f(a2)=f(a)2.
Problem
Source: 2021 USAJMO Problem 1
Tags: function, USAJMO, functional equation
15.04.2021 20:13
Claim: f2(m2+n2)=f(2mn)f(m2−n2) for all m>n. Proof: The RHS equals f((m2−n2)2+(2mn)2)=f((m2+n2)2)=f2(m2+n2) hence done ◼. Therefore f(m)=f(n)=1 implies f(2mn)=f(m2−n2)=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≥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.
15.04.2021 20:16
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\cdot1f(a^2) = f(a)^2 \Rightarrow 1 = 1^2We will prove this using strong induction: Base Case: n=1,2 f(1^2) = f(1)^2f(1)\left(f(1) - 1\right) = 0However, 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)^2However, 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)^2This 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)^2Since k<n, by inductive hypothesis: f(n)f(k^2-1)=1However, 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)^2f(2k(k+1))f(n)=f(k)^2f(k+1)^2Again, since k,k+1<n, by inductive hypothesis: f(2k(k+1))f(n) = 1Again, 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.
15.04.2021 20:27
15.04.2021 20:31
Was this possible without splitting into even/odd? I'm did something weird and now I don't know if it's legit.
15.04.2021 20:36
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.
15.04.2021 20:58
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
15.04.2021 21:05
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.
15.04.2021 21:09
vvluo wrote: Way too hard for a J1 imo you can say that again i wasted the whole 4.5 hours on this
15.04.2021 21:23
Very contrived . 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.
15.04.2021 21:26
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
15.04.2021 22:19
bump plsss sry idk my solution looks different so im thinking there is some error somewhere and idk where
15.04.2021 22:22
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)^2f\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.
15.04.2021 22:25
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.
15.04.2021 22:28
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
15.04.2021 22:29
am I the only one that substituted a bunch of numbers and variables and don't know how to continue?
15.04.2021 22:37
tastymath75025 wrote: My problem What did you write it for originally? Was it realized during the testsolving process that f(a^2)=f(a)^2 isn't necessary (assuming my proof is right).
15.04.2021 22:43
i3435 wrote: tastymath75025 wrote: My problem What did you write it for originally? Was it realized during the testsolving process that f(a^2)=f(a)^2 isn't necessary (assuming my proof is right). It made the problem easier. Still unsuitable for j1 imo.
15.04.2021 23:36
i3435 wrote: tastymath75025 wrote: My problem What did you write it for originally? Was it realized during the testsolving process that f(a^2)=f(a)^2 isn't necessary (assuming my proof is right). yeah you're right, the condition is unnecessary, however I wouldn't have solved it without that condition - I wouldn't have realized there was a pythagorean property. Without this condition it probably would have been JMO 3/6
16.04.2021 01:17
SKeole wrote: i3435 wrote: tastymath75025 wrote: My problem What did you write it for originally? Was it realized during the testsolving process that f(a^2)=f(a)^2 isn't necessary (assuming my proof is right). yeah you're right, the condition is unnecessary, however I wouldn't have solved it without that condition - I wouldn't have realized there was a pythagorean property. Without this condition it probably would have been JMO 3/6 IMO (in my opinion) this is already harder than JMO 3
02.08.2023 01:36
mathmax12 wrote: we can proove by induction that f(x)=1 in the only solution okay dont put a sketch after so many solutions have been written. Honestly, not that bad for a P1, very well suited. First note that f(1)=1,f(2)=1. We proceed by strong induction with inductive hypothesis that all integers k<n we have f(k)=1. The reason for this is obvious, but the next few steps are well motivated because taking some c^2=a^2+b^2\implies f(c)^2=f(a^2+b^2)=1, which means all (a,b,c)=(2mn,m^2-n^2,m^2+n^2) help a lot. (This decomposition of taking some f(c^2)=f((m^2-n^2)^2+(2mn)^2)=f(m^2-n^2)f(2mn) will come in useful later, and this simplification is omitted below!) Suppose n=2m. Notice f(m^2+1)=f(m)f(1)=1\implies 1=f(m^2+1)^2=f((m^2+1)^2)=f(m^2-1)f(2m)\implies f(n)=1,which implies the strong induction. \square On the other hand, if n=2m+1, notice f(m^2+(m+1)^2)=f(m)f(m+1)=1\implies 1=f(m^2+(m+1)^2)^2=f((m^2+(m+1)^2)^2)=f(2((m+1)m)^2)f((m+1)^2-m^2)\implies f(n)=1,which finishes. \blacksquare
20.08.2023 04:40
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
26.09.2023 19:50
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.
17.10.2023 09:03
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.
17.12.2023 23:56
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.
31.12.2023 11:09
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.
24.01.2024 06:24
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)^2and if n=2k+1 is odd, f(2k+1)f(2k^2+2k)=f(2k^2+2k+1)^2=f(k)^2f(k+1)^2and we are done by induction.
13.03.2024 18:35
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
13.03.2024 23:58
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.
02.04.2024 00:16
24.04.2024 22:49
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.
26.04.2024 00:44
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
15.08.2024 19:32
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)^2which 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.
22.08.2024 07:53
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.
09.09.2024 13:57
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}