2012 IMO Problems/Problem 4: Difference between revisions

Latest revision as of 07:08, 20 November 2023

Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a, b,$ and $c$ that satisfy $a + b+ c = 0$ , the following equality holds: $f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).$ (Here $\mathbb{Z}$ denotes the set of integers.)

Solution

Consider $a = b = c = 0.$ Then $f(0)^2 + f(0)^2 + f(0)^2 = 2f(0)f(0) + 2f(0)f(0) + 2f(0)f(0) \Rightarrow 3f(0)^2 = 6f(0)^2 \Rightarrow$ $f(0) = 0.$ Now we look at $b = -a, c = 0.$ $f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \Rightarrow$ $f(a)^2 + f(-a)^2 = 2f(a)f(-a) \Rightarrow$ $f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \Rightarrow$ $(f(a) - f(-a))^2=0 \Rightarrow$ $f(a)=f(-a).$

We can write $f(c)^2 - 2f(c)(f(a)+f(b)) + (f(a)-f(b))^2 = 0 \Rightarrow$

$f(c) = f(-c) = f(a+b) =\frac{2(f(a)+f(b)) \pm \sqrt{4(f(a)+f(b))^2 - 4(f(a)-f(b))^2}}{2}$

$\Rightarrow f(a+b) = f(a) + f(b) \pm 2\sqrt{f(a)f(b)}$

If $f(b) = 0$ , then $f(a+b) = f(a) = f((a)mod(b))$

Case 1: $f(1) = 0 \Rightarrow f(x)= 0$ $\forall$ $x.$ $\Box$

Case 2: $f(1) \not= 0$ , we will have $f(2) = f(1) + f(1) \pm 2\sqrt{f(1)f(1)} \Rightarrow f(2) = 0$ or $f(2) = 4f(1)$

Case 2.1: $f(1) \not= 0, f(2) = 0 \Rightarrow f(x) = f(x$ $mod$ $2) \Rightarrow f(x) = f(1)$ if $x$ is odd, $f(x) = 0$ if $x$ is even. $\Box$

Case 2.2: $f(1) \not= 0, f(2) = 4f(1) \Rightarrow f(3) = f(2) + f(1) \pm 2\sqrt{f(2)f(1)}$

$\Rightarrow f(3) = 5f(1) \pm 4f(1) \Rightarrow f(3) = f(1)$ or $9f(1)$

Case 2.2.1: $f(1) \not= 0, f(2) = 4f(1), f(3) = f(1).$

$\Rightarrow f(4) = f(1) + f(3) \pm 2\sqrt{f(1)f(3)}$ and $f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)}$

$\Rightarrow f(4) = f(1)$ or $0$ and $f(4) = 16f(1)$ or $0$

$\Rightarrow f(4) = 0 \Rightarrow f(x) = f(x$ $mod$ $4).\Box$

Case 2.2.2: $f(1) \not= 0, f(2) = 4f(1), f(3) = 9f(1).$

$\Rightarrow f(4) = f(1 + 3) = f(1) + f(3) \pm 2\sqrt{f(1)f(3)} = 16f(1)$ or $4f(1)$

and $f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)} = 16f(1)$ or $0. \Rightarrow f(4) = 16f(1).$

If $x \le 4$ then $f(x) = f(1)x^2.$

We will prove by induction $f(x) = f(1)x^2$ $\forall$ $x.$

If $x \le m$ then $f(x) = f(1)x^2.$ $\forall$ $x$ is true for some $m$ .

and if the statement is true for $m=k$

$\Rightarrow f(k+1) = f(k) + f(1) \pm 2\sqrt{f(k)f(1)} = f(1)(k+1)^2$ or $f(1)(k-1)^2$

and $f(k+1) = f(k-1) + f(2) \pm 2\sqrt{f(k-1)f(2)} = f(1)(k+1)^2$ or $f(1)(k-3)^2$

$\Rightarrow f(k+1) = f(1)(k+1)^2.$

$\Rightarrow$ the statement is true for $m=k+1$ as well.

As the statement is true for $m = 4$ , by mathematical induction we can conclude

$f(x) = f(1)x^2$ $\forall$ $x.\Box$

So, Case 2.1, Case 2.2.1 and Case 2.2.2 are the three independent possible solutions.

--Dineshram

@@ Line 6: / Line 6: @@
 Consider <math>a = b = c = 0.</math> Then <math>f(0)^2 + f(0)^2 + f(0)^2 = 2f(0)f(0) + 2f(0)f(0) + 2f(0)f(0) \Rightarrow 3f(0)^2 = 6f(0)^2 \Rightarrow</math> <cmath>f(0) = 0.</cmath>
 Now we look at <math>b = -a, c = 0.</math> <math>f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \Rightarrow</math> <math>f(a)^2 + f(-a)^2 = 2f(a)f(-a) \Rightarrow</math> <math>f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \Rightarrow</math> <math>(f(a) - f(-a))^2=0 \Rightarrow</math> <cmath>f(a)=f(-a).</cmath>
-What about <math>b = a, c = -2a?</math> Then <math>f(a)^2 + f(a)^2 + f(-2a)^2 = 2f(a)f(a) + 2f(a)f(-2a) + 2f(-2a)f(a) \Rightarrow</math> <math>2f(a)^2 + f(-2a)^2 = 2f(a)^2 + 4f(a)f(-2a) \Rightarrow</math> <math>f(-2a)^2 = 4f(a)f(-2a) \Rightarrow</math> <cmath>f(-2a) = f(2a) = 4f(a).</cmath>
-We conjecture that <math>f(na) = n^2f(a).</math> Consider <math>b = n\cdot{a}, c = -(n+1)\cdot{a}</math> and assume that <math>f(a) \neq 0.</math> If it does, we get that the constant 0 function satisfies the conditions of the problem.
-<math>f(a)^2 + f(n\cdot{a})^2 + f((n+1)\cdot{a})^2 =</math> <math>2f(a)f(n\cdot{a}) + 2f(n\cdot{a})f((n+1)\cdot{a}) + 2f((n+1)\cdot{a})f(a) \Rightarrow</math>
-<math>f(a)^2 + (n^2f(a))^2 + ((n+1)^2f(a))^2 =</math> <math>2n^2f(a)^2 + 2n^2(n+1)^2f(a)^2 + 2(n+1)^2f(a)^2 \Rightarrow</math>
-<math>1 + n^4 + (n+1)^4 = 2n^2 + 2n^2(n+1)^2 +  2(n+1)^2 \Rightarrow</math>
-<math>1 + n^4 + n^4 + 4n^3 + 6n^2 + 4n + 1 = 2n^2 + 2n^4 + 4n^3 + 2n^2 + 2n^2 + 4n +2 \Rightarrow</math>
-<math>2n^4 + 4n^3 + 6n^2 + 4n +2 = 2n^4 + 4n^3 + 6n^2 + 4n + 2 \hspace{7 mm} \checkmark</math>
-<cmath>f(n\cdot{a}) = n^2f(a)</cmath>
-We note that <math>f(a)=a^2f(1).</math> This means that we want to find what the possible values of <math>f(1)</math> are in order to finish the problem. <math>a + b + c = 0 \Rightarrow</math> <math>c = -(a+b) \Rightarrow</math> <math>a^4f(1)^2 + b^4f(1)^2 + (a+b)^4f(1)^2 =</math> <math>2a^2b^2f(1)^2 + 2b^2(a+b)^2f(1)^2 + 2(a+b)^2a^2f(1)^2.</math> Returning to the above manipulations (the ones used to show that <math>f(n\cdot{a}) = n^2f(a)</math>), we see that letting <math>n = \frac{a}{b}</math> and multiplying through by <math>b^4</math> yields precisely this result (again, assuming that <math>f(1) \neq 0,</math> with equality yielding the fact that the constant 0 function satisfies the condition). Therefore, <math>f(1)</math> can be any integer value (since <math>f</math> maps the integers to the integers only), and setting <math>f(1) = m</math>, we see that any function of the form <cmath>f(a) = m\cdot{a^2}, m \in \mathbb{Z}</cmath> satisfies the condition.
--Gadmaget
+We can write <math>f(c)^2 - 2f(c)(f(a)+f(b)) + (f(a)-f(b))^2 = 0  \Rightarrow</math>
+<cmath>f(c) = f(-c) = f(a+b) =\frac{2(f(a)+f(b)) \pm \sqrt{4(f(a)+f(b))^2 - 4(f(a)-f(b))^2}}{2}</cmath>
+<cmath>\Rightarrow f(a+b) = f(a) + f(b) \pm 2\sqrt{f(a)f(b)}</cmath>
+If <math>f(b) = 0</math>, then <cmath>f(a+b) = f(a) = f((a)mod(b))</cmath>
+'''Case 1''': <math>f(1) = 0  \Rightarrow f(x)= 0</math> <math>\forall</math> <math>x.</math> <math>\Box</math>
+Case 2: <math>f(1) \not= 0</math>, we will have <math>f(2) = f(1) + f(1) \pm 2\sqrt{f(1)f(1)} \Rightarrow f(2) = 0</math> or <math>f(2) = 4f(1)</math>
+'''Case 2.1''': <math>f(1) \not= 0, f(2) = 0 \Rightarrow f(x) = f(x</math> <math>mod</math> <math>2) \Rightarrow f(x) = f(1)</math> if <math>x</math> is odd, <math>f(x) = 0</math> if <math>x</math> is even. <math>\Box</math>
+Case 2.2: <math>f(1) \not= 0, f(2) = 4f(1) \Rightarrow f(3) = f(2) + f(1) \pm 2\sqrt{f(2)f(1)}</math>
+<math> \Rightarrow f(3) = 5f(1) \pm 4f(1) \Rightarrow f(3) = f(1)</math> or <math>9f(1)</math>
+'''Case 2.2.1''': <math>f(1) \not= 0, f(2) = 4f(1), f(3) = f(1).</math>
+<math>\Rightarrow f(4) = f(1) + f(3) \pm 2\sqrt{f(1)f(3)}</math> and <math>f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)}</math>
+<math>\Rightarrow f(4) = f(1)</math> or <math>0</math> and <math>f(4) = 16f(1)</math> or <math>0</math>
+<math>\Rightarrow f(4) = 0 \Rightarrow  f(x) = f(x</math> <math>mod</math> <math>4).\Box</math>
+'''Case 2.2.2''': <math>f(1) \not= 0, f(2) = 4f(1), f(3) = 9f(1).</math>
+<math>\Rightarrow f(4) = f(1 + 3) =  f(1) + f(3) \pm 2\sqrt{f(1)f(3)} = 16f(1)</math> or <math>4f(1)</math>
+and <math>f(4) = f(2) + f(2) \pm 2\sqrt{f(2)f(2)} = 16f(1)</math> or <math>0. \Rightarrow f(4) = 16f(1).</math>
+If <math>x \le 4</math> then <math>f(x) = f(1)x^2.</math>
+We will prove by induction <math>f(x) = f(1)x^2</math> <math>\forall</math> <math>x.</math>
+If <math>x \le m</math> then <math>f(x) = f(1)x^2.</math> <math>\forall</math> <math>x</math> is true for some <math>m</math>.
+and if the statement is true for <math>m=k</math>
+<math>\Rightarrow f(k+1) =  f(k) + f(1) \pm 2\sqrt{f(k)f(1)} = f(1)(k+1)^2</math> or <math>f(1)(k-1)^2</math>
+and <math>f(k+1) =  f(k-1) + f(2) \pm 2\sqrt{f(k-1)f(2)} = f(1)(k+1)^2</math> or <math>f(1)(k-3)^2</math>
+<math>\Rightarrow f(k+1) = f(1)(k+1)^2.</math>
+<math>\Rightarrow</math> the statement is true for <math>m=k+1</math> as well.
+As the statement is true for <math>m = 4</math>, by mathematical induction we can conclude
+<math>f(x) = f(1)x^2</math> <math>\forall</math> <math>x.\Box</math>
+'''So, Case 2.1, Case 2.2.1 and Case 2.2.2 are the three independent possible solutions.'''
+--Dineshram
+==See Also==
+*[[IMO Problems and Solutions]]
+{{IMO box|year=2012|num-b=3|num-a=5}}
+[[Category:Olympiad Algebra Problems]]
+[[Category:Functional Equation Problems]]

2012 IMO (Problems) • Resources
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2012 IMO Problems/Problem 4: Difference between revisions

Latest revision as of 07:08, 20 November 2023

Solution

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