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Indian National Mathematical Olympiad

Algebra Difficulty 6.7 National Olympiad Prove it India

Problem:

Let f:Z→Zf: \mathbb{Z} \rightarrow \mathbb{Z} be a function satisfying f(0)≠0f(0) \neq 0, f(1)=0f(1)=0 and

(i) f(xy)+f(x)f(y)=f(x)+f(y)f(xy)+f(x)f(y)=f(x)+f(y);

(ii) (f(x−y)−f(0))f(x)f(y)=0(f(x-y)-f(0)) f(x) f(y)=0,

for all x,y∈Zx, y \in \mathbb{Z}, simultaneously.

a. Find the set of all possible values of the function ff.

b. If f(10)≠0f(10) \neq 0 and f(2)=0f(2)=0, find the set of all integers nn such that f(n)≠0f(n) \neq 0.

Solution

Solution:

Setting y=0y=0 in the condition (ii), we get
(f(x)−f(0))f(x)=0 (f(x)-f(0)) f(x)=0
for all xx (since f(0)≠0f(0) \neq 0). Thus either f(x)=0f(x)=0 or f(x)=f(0)f(x)=f(0), for all x∈Zx \in \mathbb{Z}. Now taking x=y=0x=y=0 in (i), we see that f(0)+f(0)2=2f(0)f(0)+f(0)^2=2 f(0). This shows that f(0)=0f(0)=0 or f(0)=1f(0)=1. Since f(0)≠0f(0) \neq 0, we must have f(0)=1f(0)=1. We conclude that
either f(x)=0 or f(x)=1 for each x∈Z \text{either } f(x)=0 \text{ or } f(x)=1 \text{ for each } x \in \mathbb{Z}
This shows that the set of all possible values of f(x)f(x) is {0,1}\{0,1\}. This completes (a).

Let S={n∈Z∣f(n)≠0}S=\{n \in \mathbb{Z} \mid f(n) \neq 0\}. Hence we must have S={n∈Z∣f(n)=1}S=\{n \in \mathbb{Z} \mid f(n)=1\} by (a). Since f(1)=0f(1)=0, 11 is not in SS. And f(0)=1f(0)=1 implies that 0∈S0 \in S. Take any x∈Zx \in \mathbb{Z} and y∈Sy \in S. Using (ii), we get
f(xy)+f(x)=f(x)+1 f(xy)+f(x)=f(x)+1
This shows that xy∈Sxy \in S. If x∈Zx \in \mathbb{Z} and y∈Zy \in \mathbb{Z} are such that xy∈Sxy \in S, then (ii) gives
1+f(x)f(y)=f(x)+f(y) 1+f(x)f(y)=f(x)+f(y)
Thus (f(x)−1)(f(y)−1)=0(f(x)-1)(f(y)-1)=0. It follows that f(x)=1f(x)=1 or f(y)=1f(y)=1; i.e., either x∈Sx \in S or y∈Sy \in S. We also observe from (ii) that x∈Sx \in S and y∈Sy \in S implies that f(x−y)=1f(x-y)=1 so that x−y∈Sx-y \in S. Thus SS has the properties:

(A) x∈Zx \in \mathbb{Z} and y∈Sy \in S implies xy∈Sxy \in S;

(B) x,y∈Zx, y \in \mathbb{Z} and xy∈Sxy \in S implies x∈Sx \in S or y∈Sy \in S;

(C) x,y∈Sx, y \in S implies x−y∈Sx-y \in S.

Now we know that f(10)≠0f(10) \neq 0 and f(2)=0f(2)=0. Hence f(10)=1f(10)=1 and 10∈S10 \in S; and 2∉S2 \notin S. Writing 10=2×510=2 \times 5 and using (B), we conclude that 5∈S5 \in S and f(5)=1f(5)=1. Hence f(5k)=1f(5k)=1 for all k∈Zk \in \mathbb{Z} by (A).

Suppose f(5k+l)=1f(5k+l)=1 for some ll, 1≤l≤41 \leq l \leq 4. Then 5k+l∈S5k+l \in S. Choose u∈Zu \in \mathbb{Z} such that lu≡1(mod5)lu \equiv 1 \pmod{5}. We have (5k+l)u∈S(5k+l)u \in S by (A). Moreover, lu=1+5mlu=1+5m for some m∈Zm \in \mathbb{Z} and
(5k+l)u=5ku+lu=5ku+5m+1=5(ku+m)+1 (5k+l)u=5ku+lu=5ku+5m+1=5(ku+m)+1
This shows that 5(ku+m)+1∈S5(ku+m)+1 \in S. However, we know that 5(ku+m)∈S5(ku+m) \in S. By (C), 1∈S1 \in S which is a contradiction. We conclude that 5k+l∉S5k+l \notin S for any ll, 1≤l≤41 \leq l \leq 4. Thus
S={5k∣k∈Z} S=\{5k \mid k \in \mathbb{Z}\}

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