APMO/md/en-apmo2002_sol.md

XIV APMO: Solutions and Marking Schemes

1. Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer.

Let

$$A_n=\frac{a_1+a_2+\cdots +a_n}{n}A\_\{n\}=\backslash frac\{a\_\{1\}+a\_\{2\}+\backslash cdots+a\_\{n\}\}\{n\}$$

Prove that

$$a_1!a_2!\dots a_n!\ge {(\lfloor A_n\rfloor !)}^{n}a\_\{1\}!a\_\{2\}!\backslash ldots\; a\_\{n\}!\backslash geq\backslash left(\backslash left\backslash lfloor\; A\_\{n\}\backslash right\backslash rfloor!\backslash right)^\{n\}$$

where $\left\lfloor A_{n}\right\rfloor$ is the greatest integer less than or equal to $A_{n}$, and $a!=1 \times 2 \times \cdots \times a$ for $a \geq 1$ (and $0!=1$ ). When does equality hold?

Solution 1.

Assume without loss of generality that $a_{1} \geq a_{2} \geq \cdots \geq a_{n} \geq 0$, and let $s=\left\lfloor A_{n}\right\rfloor$. Let $k$ be any (fixed) index for which $a_{k} \geq s \geq a_{k+1}$.

Our inequality is equivalent to proving that

$$\frac{a_1!}{s!}\cdot \frac{a_2!}{s!}\cdot \dots \cdot \frac{a_k!}{s!}\ge \frac{s!}{a_{k+1}!}\cdot \frac{s!}{a_{k+2}!}\cdot \dots \cdot \frac{s!}{a_n!}\backslash frac\{a\_\{1\}!\}\{s!\}\; \backslash cdot\; \backslash frac\{a\_\{2\}!\}\{s!\}\; \backslash cdot\; \backslash ldots\; \backslash cdot\; \backslash frac\{a\_\{k\}!\}\{s!\}\; \backslash geq\; \backslash frac\{s!\}\{a\_\{k+1\}!\}\; \backslash cdot\; \backslash frac\{s!\}\{a\_\{k+2\}!\}\; \backslash cdot\; \backslash ldots\; \backslash cdot\; \backslash frac\{s!\}\{a\_\{n\}!\}$$

Now for $i=1,2, \ldots, k, a_{i}!/ s!$ is the product of $a_{i}-s$ factors. For example, 9!/5! $=9 \cdot 8 \cdot 7 \cdot 6$. The left side of inequality (1) therefore is the product of $A=a_{1}+a_{2}+\cdots+a_{k}-k s$ factors, all of which are greater than $s$. Similarly, the right side of (1) is the product of $B=(n-k) s-$ $\left(a_{k+1}+a_{k+2}+\cdots+a_{n}\right)$ factors, all of which are at most $s$. Since $\sum_{i=1}^{n} a_{i}=n A_{n} \geq n s, A \geq B$. This proves the inequality. [ 5 marks to here.]

Equality in (1) holds if and only if either: (i) $A=B=0$, that is, both sides of (1) are the empty product, which occurs if and only if $a_{1}=a_{2}=\cdots=a_{n}$; or (ii) $a_{1}=1$ and $s=0$, that is, the only factors on either side of (1) are 1 's, which occurs if and only if $a_{i} \in{0,1}$ for all $i$. [2 marks for both (i) and (ii), no marks for (i) only.]

Solution 2. Assume without loss of generality that $0 \leq a_{1} \leq a_{2} \leq \cdots \leq a_{n}$. Let $d=a_{n}-a_{1}$ and $m=\left|\left{i: a_{i}=a_{1}\right}\right|$. Our proof is by induction on $d$.

We first do the case $d=a_{n}-a_{1}=0$ or 1 separately. Then $a_{1}=a_{2}=\cdots=a_{m}=a$ and $a_{m+1}=\cdots=a_{n}=a+1$ for some $1 \leq m \leq n$ and $a \geq 0$. In this case we have $\left\lfloor A_{n}\right\rfloor=a$, so the inequality to be proven is just $a_{1}!a_{2}!\ldots a_{n}!\geq(a!)^{n}$, which is obvious. Equality holds if and only if either $m=n$, that is, $a_{1}=a_{2}=\cdots=a_{n}=a$; or if $a=0$, that is, $a_{1}=\cdots=a_{m}=0$ and $a_{m+1}=\cdots=a_{n}=1$. [ 2 marks to here.]

So assume that $d=a_{n}-a_{1} \geq 2$ and that the inequality holds for all sequences with smaller values of $d$, or with the same value of $d$ and smaller values of $m$. Then the sequence

$$a_1+1,a_2,a_3,\dots ,a_{n-1},a_n-1a\_\{1\}+1,\; a\_\{2\},\; a\_\{3\},\; \backslash ldots,\; a\_\{n-1\},\; a\_\{n\}-1$$

though not necessarily in non-decreasing order any more, does have either a smaller value of $d$, or the same value of $d$ and a smaller value of $m$, but in any case has the same value of $A_{n}$. Thus, by induction and since $a_{n}>a_{1}+1$,

which completes the proof. Equality cannot hold in this case. 2. Find all positive integers $a$ and $b$ such that

$$\frac{a^2+b}{b^2-a}\text{ and }\frac{b^2+a}{a^2-b}\backslash frac\{a^\{2\}+b\}\{b^\{2\}-a\}\; \backslash text\; \{\; and\; \}\; \backslash frac\{b^\{2\}+a\}\{a^\{2\}-b\}$$

are both integers. Solution. By the symmetry of the problem, we may suppose that $a \leq b$. Notice that $b^{2}-a \geq 0$, so that if $\frac{a^{2}+b}{b^{2}-a}$ is a positive integer, then $a^{2}+b \geq b^{2}-a$. Rearranging this inequality and factorizing, we find that $(a+b)(a-b+1) \geq 0$. Since $a, b>0$, we must have $a \geq b-1$. [3 marks to here.] We therefore have two cases:

Case 1: $a=b$. Substituting, we have

$$\frac{a^2+a}{a^2-a}=\frac{a+1}{a-1}=1+\frac{2}{a-1},\backslash frac\{a^\{2\}+a\}\{a^\{2\}-a\}=\backslash frac\{a+1\}\{a-1\}=1+\backslash frac\{2\}\{a-1\},$$

which is an integer if and only if $(a-1) \mid 2$. As $a>0$, the only possible values are $a-1=1$ or 2. Hence, $(a, b)=(2,2)$ or $(3,3)$. [1 mark.]

Case 2: $a=b-1$. Substituting, we have

$$\frac{b^2+a}{a^2-b}=\frac{(a+1{)}^2+a}{a^2-(a+1)}=\frac{a^2+3a+1}{a^2-a-1}=1+\frac{4a+2}{a^2-a-1}\mathrm{.}\backslash frac\{b^\{2\}+a\}\{a^\{2\}-b\}=\backslash frac\{(a+1)^\{2\}+a\}\{a^\{2\}-(a+1)\}=\backslash frac\{a^\{2\}+3\; a+1\}\{a^\{2\}-a-1\}=1+\backslash frac\{4\; a+2\}\{a^\{2\}-a-1\}\; .$$

Once again, notice that $4 a+2>0$, and hence, for $\frac{4 a+2}{a^{2}-a-1}$ to be an integer, we must have $4 a+2 \geq a^{2}-a-1$, that is, $a^{2}-5 a-3 \leq 0$. Hence, since $a$ is an integer, we can bound $a$ by $1 \leq \bar{a} \leq 5$. Checking all the ordered pairs $(a, b)=(1,2),(2,3), \ldots,(5,6)$, we find that only $(1,2)$ and $(2,3)$ satisfy the given conditions. [3 marks.]

Thus, the ordered pairs that work are

$$(2,2),(3,3),(1,2),(2,3),(2,1),(3,2)(2,2),(3,3),(1,2),(2,3),(2,1),(3,2)$$

where the last two pairs follow by symmetry. [2 marks if these solutions are found without proof that there are no others.] 3. Let $A B C$ be an equilateral triangle. Let $P$ be a point on the side $A C$ and $Q$ be a point on the side $A B$ so that both triangles $A B P$ and $A C Q$ are acute. Let $R$ be the orthocentre of triangle $A B P$ and $S$ be the orthocentre of triangle $A C Q$. Let $T$ be the point common to the segments $B P$ and $C Q$. Find all possible values of $\angle C B P$ and $\angle B C Q$ such that triangle $T R S$ is equilateral.

Solution.

We are going to show that this can only happen when

$$\mathrm{\angle }CBP=\mathrm{\angle }BCQ={15}^{\circ }\mathrm{.}\backslash angle\; C\; B\; P=\backslash angle\; B\; C\; Q=15^\{\backslash circ\}\; .$$

Lemma. If $\angle C B P>\angle B C Q$, then $R T>S T$. Proof. Let $A D, B E$ and $C F$ be the altitudes of triangle $A B C$ concurrent at its centre $G$. Then $P$ lies on $C E, Q$ lies on $B F$, and thus $T$ lies in triangle $B D G$.

Note that -

$$\mathrm{\angle }FAS=\mathrm{\angle }FCQ={30}^{\circ }-\mathrm{\angle }BCQ>{30}^{\circ }-\mathrm{\angle }CBP=\mathrm{\angle }EBP=\mathrm{\angle }EAR\backslash angle\; F\; A\; S=\backslash angle\; F\; C\; Q=30^\{\backslash circ\}-\backslash angle\; B\; C\; Q>30^\{\backslash circ\}-\backslash angle\; C\; B\; P=\backslash angle\; E\; B\; P=\backslash angle\; E\; A\; R$$

Since $A F=A E$, we have $F S>E R$ so that

Let $T_{x}$ be the projection of $T$ onto $B C$ and $T_{y}$ be the projection of $T$ onto $A D$, and similarly for $R$ and $S$. We have

$$R_x{T}_x=D{R}_x+D{T}_x>\mid D{S}_x-D{T}_x\mid =S_x{T}_{x}R\_\{x\}\; T\_\{x\}=D\; R\_\{x\}+D\; T\_\{x\}>\backslash left|D\; S\_\{x\}-D\; T\_\{x\}\backslash right|=S\_\{x\}\; T\_\{x\}$$

and

$$R_y{T}_y=G{R}_y+G{T}_y>G{S}_y+G{T}_y=S_y{T}_{y}R\_\{y\}\; T\_\{y\}=G\; R\_\{y\}+G\; T\_\{y\}>G\; S\_\{y\}+G\; T\_\{y\}=S\_\{y\}\; T\_\{y\}$$

It follows that $R T>S T$. [1 mark for stating the Lemma, 3 marks for proving it.] Thus, if $\triangle T R S$ is equilateral, we must have $\angle C B P=\angle B C Q$.

It is clear from the symmetry of the figure that $T R=T S$, so $\triangle T R S$ is equilateral if and only if $\angle R T A=30^{\circ}$. Now, as $B R$ is an altitude of the triangle $A B C, \angle R B A=30^{\circ}$. So $\triangle T R S$ is equilateral if and only if $R T B A$ is a cyclic quadrilateral. Therefore, $\triangle T R S$ is equilateral if and only if $\angle T B R=\angle T A R$. But

and so

$${30}^{\circ }=\mathrm{\angle }TAR+\mathrm{\angle }TBR30^\{\backslash circ\}=\backslash angle\; T\; A\; R+\backslash angle\; T\; B\; R$$

But these angles must be equal, so $\angle T A R=\angle T B R=15^{\circ}$. Therefore $\angle C B P=\angle B C Q=15^{\circ}$. [3 marks for finishing the proof with the assumption that $\angle C B P=\angle B C Q$.] 4. Let $x, y, z$ be positive numbers such that

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\backslash frac\{1\}\{x\}+\backslash frac\{1\}\{y\}+\backslash frac\{1\}\{z\}=1$$

Show that

$$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge \sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\backslash sqrt\{x+y\; z\}+\backslash sqrt\{y+z\; x\}+\backslash sqrt\{z+x\; y\}\; \backslash geq\; \backslash sqrt\{x\; y\; z\}+\backslash sqrt\{x\}+\backslash sqrt\{y\}+\backslash sqrt\{z\}$$

Solution 1.

Note. It is easy to check that equality holds if and only if $x=y=z=3$. Solution 2. Squaring both sides of the given inequality, we obtain

It follows from the given condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ that $x y z=\sum_{\text {cyclic }} x y$. Therefore, the given inequality is equivalent to

$$\sum _{\text{cyclic }}\sqrt{x+yz}\sqrt{y+zx}\ge \sqrt{xyz}\sum _{\text{cyclic }}\sqrt{x}+\sum _{\text{cyclic }}\sqrt{xy}\mathrm{.}[2\text{ marks.] }\backslash sum\_\{\backslash text\; \{cyclic\; \}\}\; \backslash sqrt\{x+y\; z\}\; \backslash sqrt\{y+z\; x\}\; \backslash geq\; \backslash sqrt\{x\; y\; z\}\; \backslash sum\_\{\backslash text\; \{cyclic\; \}\}\; \backslash sqrt\{x\}+\backslash sum\_\{\backslash text\; \{cyclic\; \}\}\; \backslash sqrt\{x\; y\}\; .\; \backslash quad[2\; \backslash text\; \{\; marks.]\; \}$$

Using the Cauchy-Schwarz inequality [or just $x^{2}+y^{2} \geq 2 x y$ ], we see that

$$(x+yz)(y+zx)\ge {(\sqrt{xy}+\sqrt{xy{z}^2})}^2,[1\text{ mark. }](x+y\; z)(y+z\; x)\; \backslash geq\backslash left(\backslash sqrt\{x\; y\}+\backslash sqrt\{x\; y\; z^\{2\}\}\backslash right)^\{2\},\; \backslash quad[1\; \backslash text\; \{\; mark.\; \}]$$

or

$$\sqrt{x+yz}\sqrt{y+zx}\ge \sqrt{xy}+\sqrt{z}\sqrt{xyz}\mathrm{.}[1\text{ mark. }]\backslash sqrt\{x+y\; z\}\; \backslash sqrt\{y+z\; x\}\; \backslash geq\; \backslash sqrt\{x\; y\}+\backslash sqrt\{z\}\; \backslash sqrt\{x\; y\; z\}\; .\; \backslash quad[1\; \backslash text\; \{\; mark.\; \}]$$

Taking the cyclic sum of this inequality over $x, y$ and $z$, we get the desired inequality. [2 marks.]

Solution 3. This is another way of presenting the idea in the first solution. Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ and the AM-GM inequality, we have

which gives

$$\sqrt{x+yz}\ge \sqrt{\frac{yz}{x}}+\sqrt{x}\mathrm{.}[3\text{ marks. }]\backslash sqrt\{x+y\; z\}\; \backslash geq\; \backslash sqrt\{\backslash frac\{y\; z\}\{x\}\}+\backslash sqrt\{x\}\; .\; \backslash quad[3\; \backslash text\; \{\; marks.\; \}]$$

Similarly, we have

$$\sqrt{y+zx}\ge \sqrt{\frac{zx}{y}}+\sqrt{y}\text{ and }\sqrt{z+xy}\ge \sqrt{\frac{xy}{z}}+\sqrt{z}\backslash sqrt\{y+z\; x\}\; \backslash geq\; \backslash sqrt\{\backslash frac\{z\; x\}\{y\}\}+\backslash sqrt\{y\}\; \backslash text\; \{\; and\; \}\; \backslash sqrt\{z+x\; y\}\; \backslash geq\; \backslash sqrt\{\backslash frac\{x\; y\}\{z\}\}+\backslash sqrt\{z\}$$

Addition yields

$$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge \sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}+\sqrt{x}+\sqrt{y}+\sqrt{z}\backslash sqrt\{x+y\; z\}+\backslash sqrt\{y+z\; x\}+\backslash sqrt\{z+x\; y\}\; \backslash geq\; \backslash sqrt\{\backslash frac\{y\; z\}\{x\}\}+\backslash sqrt\{\backslash frac\{z\; x\}\{y\}\}+\backslash sqrt\{\backslash frac\{x\; y\}\{z\}\}+\backslash sqrt\{x\}+\backslash sqrt\{y\}+\backslash sqrt\{z\}$$

[2 marks.] Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ again, we have

$$\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}=\sqrt{xyz}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\sqrt{xyz},[1\text{ mark. }]\backslash sqrt\{\backslash frac\{y\; z\}\{x\}\}+\backslash sqrt\{\backslash frac\{z\; x\}\{y\}\}+\backslash sqrt\{\backslash frac\{x\; y\}\{z\}\}=\backslash sqrt\{x\; y\; z\}\backslash left(\backslash frac\{1\}\{x\}+\backslash frac\{1\}\{y\}+\backslash frac\{1\}\{z\}\backslash right)=\backslash sqrt\{x\; y\; z\},\; \backslash quad[1\; \backslash text\; \{\; mark.\; \}]$$

and thus

$$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge \sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\mathrm{.}[1\text{ mark. }]\backslash sqrt\{x+y\; z\}+\backslash sqrt\{y+z\; x\}+\backslash sqrt\{z+x\; y\}\; \backslash geq\; \backslash sqrt\{x\; y\; z\}+\backslash sqrt\{x\}+\backslash sqrt\{y\}+\backslash sqrt\{z\}\; .\; \backslash quad[1\; \backslash text\; \{\; mark.\; \}]$$

Solution 4.

This is also another way of presenting the idea in the first solution. We make the substitution $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$. Then it is enough to show that

$$\sqrt{\frac{1}{a}+\frac{1}{bc}}+\sqrt{\frac{1}{b}+\frac{1}{ca}}+\sqrt{\frac{1}{c}+\frac{1}{ab}}\ge \sqrt{\frac{1}{abc}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}},\backslash sqrt\{\backslash frac\{1\}\{a\}+\backslash frac\{1\}\{b\; c\}\}+\backslash sqrt\{\backslash frac\{1\}\{b\}+\backslash frac\{1\}\{c\; a\}\}+\backslash sqrt\{\backslash frac\{1\}\{c\}+\backslash frac\{1\}\{a\; b\}\}\; \backslash geq\; \backslash sqrt\{\backslash frac\{1\}\{a\; b\; c\}\}+\backslash sqrt\{\backslash frac\{1\}\{a\}\}+\backslash sqrt\{\backslash frac\{1\}\{b\}\}+\backslash sqrt\{\backslash frac\{1\}\{c\}\},$$

where $a+b+c=1$. Multiplying this inequality by $\sqrt{a b c}$, we find that it can be written

$$\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge 1+\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\mathrm{.}[1\text{ mark. }]\backslash sqrt\{a+b\; c\}+\backslash sqrt\{b+c\; a\}+\backslash sqrt\{c+a\; b\}\; \backslash geq\; 1+\backslash sqrt\{b\; c\}+\backslash sqrt\{c\; a\}+\backslash sqrt\{a\; b\}\; .\; \backslash quad[1\; \backslash text\; \{\; mark.\; \}]$$

This is equivalent to

which in turn is equivalent to

$$\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)}\ge a+b+c+\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\backslash sqrt\{(a+b)(a+c)\}+\backslash sqrt\{(b+c)(b+a)\}+\backslash sqrt\{(c+a)(c+b)\}\; \backslash geq\; a+b+c+\backslash sqrt\{b\; c\}+\backslash sqrt\{c\; a\}+\backslash sqrt\{a\; b\}$$

[1 mark.] (This is a homogeneous version of the original inequality.) By the Cauchy-Schwarz inequality (or since $b+c \geq 2 \sqrt{b c}$ ), we have

$$[(\sqrt{a}{)}^2+(\sqrt{b}{)}^2][(\sqrt{a}{)}^2+(\sqrt{c}{)}^2]\ge (\sqrt{a}\sqrt{a}+\sqrt{b}\sqrt{c}{)}^2\backslash left[(\backslash sqrt\{a\})^\{2\}+(\backslash sqrt\{b\})^\{2\}\backslash right]\backslash left[(\backslash sqrt\{a\})^\{2\}+(\backslash sqrt\{c\})^\{2\}\backslash right]\; \backslash geq(\backslash sqrt\{a\}\; \backslash sqrt\{a\}+\backslash sqrt\{b\}\; \backslash sqrt\{c\})^\{2\}$$

or

$$\sqrt{(a+b)(a+c)}\ge a+\sqrt{bc}\mathrm{.}[2\text{ marks. }]\backslash sqrt\{(a+b)(a+c)\}\; \backslash geq\; a+\backslash sqrt\{b\; c\}\; .\; \backslash quad[2\; \backslash text\; \{\; marks.\; \}]$$

Taking the cyclic sum of this inequality over $a, b, c$, we get the desired inequality. [2 marks.] 5. Let R denote the set of all real numbers. Find all functions $f$ from R to R satisfying: (i) there are only finitely many $s$ in R such that $f(s)=0$, and (ii) $f\left(x^{4}+y\right)=x^{3} f(x)+f(f(y))$ for all $x, y$ in $\mathbf{R}$.

Solution 1.

The only such function is the identity function on $R$. Setting $(x, y)=(1,0)$ in the given functional equation (ii), we have $f(f(0))=0$. Setting $x=0$ in (ii), we find

$$f(y)=f(f(y))f(y)=f(f(y))$$

[1 mark.] and thus $f(0)=f(f(0))=0$ [1 mark.]. It follows from (ii) that $f\left(x^{4}+y\right)=$ $x^{3} f(x)+f(y)$ for all $x, y \in \mathbf{R}$. Set $y=0$ to obtain

$$f\left(x^4\right)=x^{3}f(x)f\backslash left(x^\{4\}\backslash right)=x^\{3\}\; f(x)$$

for all $x \in \mathrm{R}$, and so

$$f(x^4+y)=f\left(x^4\right)+f(y)f\backslash left(x^\{4\}+y\backslash right)=f\backslash left(x^\{4\}\backslash right)+f(y)$$

for all $x, y \in \mathbf{R}$. The functional equation (3) suggests that $f$ is additive, that is, $f(a+b)=$ $f(a)+f(b)$ for all $a, b \in \mathbf{R}$. [1 mark.] We now show this.

First assume that $a \geq 0$ and $b \in R$. It follows from (3) that

$$f(a+b)=f({\left(a^{1\mathrm{/}4}\right)}^4+b)=f\left({\left(a^{1\mathrm{/}4}\right)}^4\right)+f(b)=f(a)+f(b)f(a+b)=f\backslash left(\backslash left(a^\{1\; /\; 4\}\backslash right)^\{4\}+b\backslash right)=f\backslash left(\backslash left(a^\{1\; /\; 4\}\backslash right)^\{4\}\backslash right)+f(b)=f(a)+f(b)$$

We next note that $f$ is an odd function, since from (2)

$$f(-x)=\frac{f\left(x^4\right)}{(-x{)}^3}=\frac{f\left(x^4\right)}{-x^3}=-f(x),x\mathrm{\ne }0f(-x)=\backslash frac\{f\backslash left(x^\{4\}\backslash right)\}\{(-x)^\{3\}\}=\backslash frac\{f\backslash left(x^\{4\}\backslash right)\}\{-x^\{3\}\}=-f(x),\; \backslash quad\; x\; \backslash neq\; 0$$

Since $f$ is odd, we have that, for $a<0$ and $b \in R$,

Therefore, we conclude that $f(a+b)=f(a)+f(b)$ for all $a, b \in R$. [2 mers.].] We now show that ${s \in \mathrm{R} \mid f(s)=0}={0}$. Recall that $f(0)=0$. Assume that there is a nonzero $h \in \mathrm{R}$ such that $f(h)=0$. Then, using the fact that $f$ is additive, we inductively have $f(n h)=0$ or $n h \in{s \in R \mid f(s)=0}$ for all $n \in \mathbf{N}$. However, this is a contradiction to the given condition (i). [1 mark.]

It's now easy to check that $f$ is one-to-one. Assume that $f(a)=f^{\prime}(b)$ for some $a, b \in \mathbb{P}$. Then, we have $f(b)=f(a)=f(a-b)+f(b)$ or $f(a-b)=0$. This implies that $a-b \in{s \in$ $\mathbf{R} \mid f(s)=0}={0}$ or $a=b$, as desired. From (1) and the fact that $f$ is one-to-one, we deduce that $f(x)=x$ for all $x \in \mathbf{R}$. [1 mark.] This completes the proof.

Solution 2.

Again, the only such function is the identity function on R . As in Solution 1, we first show that $f(f(y))=f(y), f(0)=0$, and $f\left(x^{4}\right)=x^{3} f(x)$. [2 marks.] From the latter follows

$$f(x)=0⟹f\left(x^4\right)=0f(x)=0\; \backslash Longrightarrow\; f\backslash left(x^\{4\}\backslash right)=0$$

and from condition (i) we get that $f(x)=0$ only possibly for $x \in{0,1,-1}$. [1 mark.] Next we prove

$$f(a)=b⟹f(\sqrt[4]{\mathrm{\mid }a-b\mathrm{\mid }})=0f(a)=b\; \backslash Longrightarrow\; f(\backslash sqrt[4]\{|a-b|\})=0$$

This is clear if $a=b$. If $a>b$ then

so $(a-b)^{3 / 4} f(\sqrt[4]{a-b})=0$ which means $f(\sqrt[4]{|a-b|})=0$. If $a<b$ we get similarly

and again $f(\sqrt[4]{|a-b|})=0$. [2 marks.] Thus $f(a)=b \Longrightarrow|a-b| \in{0,1}$. Suppose that $f(x)=x+b$ for some $x$, where $|b|=1$. Then from $f\left(x^{4}\right)=x^{3} f(x)$ and $f\left(x^{4}\right)=x^{4}+a$ for some $|a| \leq 1$ we get $x^{3}=a / b$, so $|x| \leq 1$. Thus $f(x)=x$ for all $x$ except possibly $x= \pm 1$. [ 1 mark.] But for example,

$$f(1)=f(2^4-15)=2^{3}f(2)+f(f(-15))=2^3\cdot 2-15=1f(1)=f\backslash left(2^\{4\}-15\backslash right)=2^\{3\}\; f(2)+f(f(-15))=2^\{3\}\; \backslash cdot\; 2-15=1$$

and

$$f(-1)=f(2^4-17)=2^{3}f(2)+f(f(-17))=2^3\cdot 2-17=-1f(-1)=f\backslash left(2^\{4\}-17\backslash right)=2^\{3\}\; f(2)+f(f(-17))=2^\{3\}\; \backslash cdot\; 2-17=-1$$

[1 mark.] This finishes the proof.