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Algebra Level 4

Denote by $S$ the set of all positive integers. A function $f: S \rightarrow S$ satisfies

$f(f^2(m) + 2f^2(n)) = m^2 + 2 n^2$

for all $m,n \in S$ .

Find $f(2016)$ .

The answer is 2016.

1 solution

Kushal Bose
Nov 10, 2016

From the pattern in the equation it may be concluded that $f(x)=x$

For confirmation I am going to show two cases.

Case(1): $f(x)<x \\ \implies f(m)<m \\ \implies f^2(m)<m^2$

Similar for $f^2(n)<n^2$

$f^2(m)+2f^2(n)<m^2+2n^2 \\ f(f^2(m)+2f^2(n)<f(m^2+2n^2)<m^2+2n^2)$

So, it contradicts the question then $f(x) \nless x$

Case(2) $f(x)>x \\ \implies f(m)>m \\ \implies f^2(m)>m^2$

Similar for $f^2(n)>n^2$

$f^2(m)+2f^2(n)>m^2+2n^2 \\ f(f^2(m)+2f^2(n)>f(m^2+2n^2)>m^2+2n^2)$

So, it contradicts the question then $f(x) \ngtr x$

So, the last option is $f(x)=x$

Your cases are assuming that $f(x) < x$ for all values of $x$ . You haven't dealt with the case where sometimes $f(x) > x$ and sometimes $f(x) =x$ and sometimes $f(x) < x$ .

This is a common misconception made when trying to work with functional equations.

Calvin Lin Staff - 4 years, 7 months ago

I will not tell the answer

The answer is 2016.

1 solution

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