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Take prime $p$ such that $p$ is co-prime to both $a,b$. Then take $n=p-1$ and use Fermat's little theorem to prove $p|a-b$. But there are infinite $p$ primes co-prime to $a-b$. implies $a=b$.

P.S.:: My solution is identical to official solution. I couldn't find any different solution. You are welcome to give another one.

Let $P(x,y)$ be the statement. $P(1,0)$ implies $f(1)=f(1)f(0)$..........(K)
Consider cases:
i)$f(0)=0$
Then, $f(1)=0$ . Now, $P(x,x-1)$ implies $f(2x-1)=f(x-1)f(x)................(*)$
Simple induction to prove $f(2n-1)=0$ for all $n\in \mathbb N$ as one of $x,x-1$ is odd.
Also $f(4n)=f(n-1)f(n+1)$................(**) .So, $f(4)=0$. We will use induction to prove $f(4n)=0$. Now note that either $n-1,n+1$ both odd or one of them is divisible by $4$. It implies $f(4n)$ must be 0..........(a).
Now, if $f(4n-2)$ is $0$ for all $n\in \mathbb N$ then $f(n)=0$ for all $n$..........(Solution-1)

Otherwise, let $f(4k-2)$ is non-zero for some $k \in \mathbb N$. Then, $P(4k-2,4k'-2)$ implies $f(4k'-2)=0$ for $k'<k$ (using (a) because here $x^2-y^2$ is divisible by 4) and $P(4k''-2,4k-2)$ implies $f(4k''-2)=0$ for $k''>k$ using (a). It implies $f(4k-2)$ is non-zero for some $k\in \mathbb N$ and $f(n)=0$ for all other $n$...........(Solution-2)

ii)$f(0)$ is non-zero but $f(1)=0$
Same way in case(i), prove that $f(2n-1)=0$, for all $n\in \mathbb N$
Now since $(2n+1)^2-(2n-1)^2=(n+2)^2-(n-2)^2$; Using $P$ we will get $f(2n+1)f(2n-1)=f(n+2)f(n-2)$........(***)
So, $f(4)=0$ using (***) and same way in case (i), $f(4n)=0$ for all $n\in \mathbb N$.
Now, $P(4n-2,0)$ implies $f((4n-2)^2)=f(4n-2)f(0)$. But it implies $f(4n-2)=0$ from previous conditions.
So $f(0)$ is non zero and $f(n)=0$ for all $n\in \mathbb N$..............(Solution-3)

iii)Both $f(0),f(1)$ are non-zero. But $f(2)=0$
Then $f(0)=1$ using $K$. Use (*) to get $f(3)=0$ and $f(5)=0$. Again use induction to get $f(2n+1)=0$ for all $n\in \mathbb N$. Use (***) setting $n=2$ to get $f(4)=0$. Again use induction to get $f(4n)=0$ for all $n\in \mathbb N$. Now $P(4n-2,0)$ implies $f((4n-2)^2)=f(4n-2)=0$ as $4|(4n-2)^2$.
So $f(0)=1$, $f(1)$ non-zero and $f(n)=0$ for all other $n\in \mathbb N$............(Solution-4)

iv)$f(0),f(1),f(2)$ are non-zero. Then $f(0)=1$.Let $f(1)=a,f(2)=b$.
Use (*) to get $f(3)=ab,f(5)=ab^2$ and use (**) to get $f(4)=a$ But using (***)
$f(4)=f(3)f(5)$ implies $a=a^2b^3$ and so $a=b=1$
Now use (*) in (***) to get $f(n+2)f(n-2)=f(n-1)f(n)^2f(n+1)$ and use induction to get $f(n)=1$ for all $n \in \mathbb N_0$..................(Solution-5)

(Done)

Check: $x^2-y^2$ is divisible by $4$ or it is an odd number (Then one of $x,y$ is odd). Now it's easy to check every solution is a perfect match.

P.S.: I'm tired with long case studies but I always try to give full solution. Many people try to skip many parts and fall into traps. I suggest everyone to give complete solution. If you find any bug , inform me. I hope for any other smart solution.

$K,L$ are the reflection of ($E$ on $BI$), ($F$ on $CI$) respectively. But $D$ is the reflection of ($E$ on $CI$) and ($F$ on $BI$). So, $\angle EDF = \angle KFD= \angle KLD$ and also $\angle FDE= \angle LED= \angle LKD$ which implies $DL=DK$ and so $DI$ is the perpendicular bisector of $KL$. So, $D,I,J$ collinear.

Let $AI$ meets $BC$ at $L$. Then $L$ is fixed. Let $W$ is the midpoint of $PQ$ and perpendicular bisector intersect $BC$ at $J$. Define $CI\cap EF =O, BI\cap EF =H$ . Angle chase will imply $\angle FOI= \frac {1}{2}\angle B$ . So, $IFOB$ cyclic. So, $BO \perp CO$. Similarly we can prove that $CH \perp BH$ . So $BC$ is the diameter of circle $BCHO$. If $Z,R$ are midpoints of $BC,HO$ respectively, then $ZR$ is perpendicular to $HO$ . But note that $(Q,N,M,W,P)$ are reflection of $(O,F,E,R,H)$ with respect to $I$. But as $ZR||IL||WZ$ and $IW=IR$, so $L$ is the midpoint of $JZ$ . As $L,Z$ are fixed, so $J$ the fixed point.

Same as SANZEED

Let $l'_1\perp DB, l'_2\perp DC$ through $H$ intersecting circle $ACH,ABH$ at $F,E$ respectively.
Let $m_1\cap m_2=A'$ where $m_1,m_2$ are perpendicular to $DB,DC$ from $C,B$ respectively.So $D$ is the orthocenter of $A'BC$. Now $A'D||AH,A'B||EH,A'C||FH$ . Also note that $\angle A'=180-\angle D= \angle A$ . Now using cyclic property, $90-\angle A = \angle AEH= \angle AFH$ . But $\angle A'BD= \angle A'CD = 90 -\angle A$
. That's why $\triangle A'CB$ and $\triangle HEF$ are homothetic.

Note that, radius of circle $AHB,AHC$ are equal (Use sine law to prove they are equal to radius of circle $ABC$)
Now, let $ME\cap NF = P'$. Then angle chase will imply $\angle MP'N= \angle A$. Again as $MP\perp EA$ and $NP \perp FA$ so, $\angle MPN= \angle A$. It implies that, $\angle EMP=\angle FNP$...(i). Also $\frac{MP}{NP}=\frac{sin\angle MNP}{sin \angle NMP}= \frac{sin \angle MHE}{ sin \angle NHF}$ (As,$HE||A'B||PN$ and $HF||A'C||PM$)
But $\frac{sin \angle MHE}{ sin \angle NHF}=\frac{ME}{NF}$ because of two equal radius.

So,$\frac{MP}{NP}=\frac{ME}{NF}$....(ii)
Using (i),(ii); a spiral similarity with center $P$ and angle $\angle A$ sends ($M,E$) to ($N,F$)
That means $\angle EPF = \angle A$. So, locus of $P$ is circle $EAF$
(Proved)