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$f$ একটা ফাংশন যার ডোমেইন ও কোডোমেইন পূর্ণসংখ্যার সেট, $\mathbb{Z}$।
$f(f(x+y))= f(x^2) + f(y^2)$
$f(f(2020)) = 1010$.
$f(2025)$-এর মান বের করো।

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$f: \mathbb{Z} \to \mathbb{Z}$
$f(f(x+y))= f(x^2) + f(y^2)$
$f(f(2020)) = 1010.$
Find $f(2025)$.

Let's substitute $y=-x$ in the first equation.
$f(f(0))=2f(x^2) \dots (1)$
Let's substitute $y=x$ in the first equation.
$f(f(2x))=2f(x^2) \dots (2)$
From $(1)$ and $(2)$, we get,
$2f(x²)=f(f(2x))=f(f(0))$, which is constant for all $x\in \mathbb{Z}$
So, $f(f(0))=f(f(2020))=1010$, by substituting $x=1010$
So, $f(2025)=\frac12\cdot 2f(45²)=\frac12\cdot f(f(0))=\frac12\cdot1010=505$