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1 (Canada 1969) Let N = {1, 2, 3, · · · } denote the set of positive integers. Find all
PEN K12 functions f : N → N such that for all m, n ∈ N: f (2) = 2, f (mn) = f (m)f (n),
f (n + 1) > f (n).
First Solution. To get some idea, we first evaluate f (n) for small positive integers n. It follows
from f (1 · 1) = f (1) · f (1) that f (1) = 1. By the multiplicity, we get f (4) = f (2)2 = 4. It
follows from the inequality 2 = f (2) < f (3) < f (4) = 4 that f (3) = 3. Also, we compute
f (6) = f (2)f (3) = 6. Since 4 = f (4) < f (5) < f (6) = 6, we get f (5) = 5.
We prove by induction that f (n) = n for all n ∈ N. We know that it holds for n = 1, 2, 3.
Now, let n > 2 and suppose that f (k) = k for all k ∈ {1, · · · , n}. We show that f (n + 1) = n + 1.
Case 1 n + 1 is composite. One may write n + 1 = ab for some positive integers a and b
with 2 ≤ a ≤ b ≤ n. By the inductive hypothesis, we have f (a) = a and f (b) = b. It follows that
f (n + 1) = f (a)f (b) = ab = n + 1.
Case 2 n + 1 is prime. In this case, n + 2 is even. Write n + 2 = 2k for some positive integer
k. Since n ≥ 2, we get 2k = n + 2 ≥ 4 or k ≥ 2. Since k =
n+2
2
≤ n, by the inductive hypothesis,
we have f (k) = k. It follows that f (n + 2) = f (2k) = f (2)f (k) = 2k = n + 2. From the inequality
n = f (n) < f (n + 1) < f (n + 2) = n + 2
(1)
we see that f (n + 1) = n + 1.
By induction, we conclude that f (n) = n for all positive integers n.
Second Solution. As in the previous solution, we get f (1) = 1. From the multiplicativity of f , we
find that f (2n) = f (2)f (n) = 2f (n) for all positive integers n. This implies that
f 2k = 2k
(2)
for all positive integers k. Let k ∈ N. From the assumption, we obtain the inequality
2k = f 2k < f 2k + 1 < · · · < f 2k+1 − 1 < f 2k+1 = 2k+1 .
(3)
In other words, the increasing sequence of 2k + 1 positive integers
f 2k , f 2k + 1 , · · · , f 2k+1 − 1 , f 2k+1
(4)
lies in the set of 2k + 1 consecutive integers {2k , 2k + 1, · · · , 2k+1 − 1, 2k+1 }. This means that
f (n) = n for all 2k ≤ n ≤ 2k+1 . Since this holds for all positive integers k, we conclude that
f (n) = n for all n ≥ 2.
Third Solution. The assumption that f (mn) = f (m)f (n) for all positive integers m and n is too
strong. We can establish the following proposition.
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Proposition 1. (Putnam 1963/A2) Let f : N → N be a strictly increasing function satisfying
that f (2) = 2 and f (mn) = f (m)f (n) for all relatively prime m and n. Then, f is the identity
function on N.
Proof Since f is strictly increasing, we find that f (n + 1) ≥ f (n) + 1 for all positive integers
n. It follows that f (n+k) ≥ f (n)+k for all positive integers n and k. We now determine p = f (3).
On the one hand, we obtain
f (18) ≥ f (15) + 3 ≥ f (3)f (5) + 3 ≥ f (3)(f (3) + 2) + 3 = p2 + 2p + 3.
(5)
On the other hand, we obtain
f (18) = f (2)f (9) ≤ 2(f (10) − 1) = 2f (2)f (5) − 2 ≤ 4(f (6) − 1) − 2 = 4f (2)f (3) − 6 = 8p − 6. (6)
Combining these two, we deduce p2 + 2p + 3 ≤ 8p − 6 or (p − 3)2 ≤ 0. So, we have f (3) = p = 3.
We now prove that f 2l + 1 = 2l + 1 for all positive integers l. Since f (3) = 3, it clearly
holds for l = 1. Assuming that f 2l + 1 = 2l + 1 for some positive integer l, we obtain
f 2l+1 + 2 = f (2)f 2l + 1 = 2 2l + 1 = 2l+1 + 2.
(7)
Since f is strictly increasing, this means that f 2l + k = 2l + k for all k ∈ {1, · · · , 2l + 2}. In
particular, we get f 2l+1 + 1 = 2l+1 + 1, as desired.
Now, we find that f (n) = n for all positive integers n. It clearly holds for n = 1, 2. Let l be
a fixed positive integer. We have f 2l + 1 = 2l + 1 and f 2l+1 + 1 = 2l+1 + 1. Since f is strictly
increasing, this means that f 2l + k = 2l + k for all k ∈ {1, · · · , 2l + 1}. Since it holds for all
positive integers l, we conclude that f (n) = n for all n ≥ 3. This completes the proof.
Fourth Solution. We can establish the following general result.
Proposition 2. Let f : N → R+ be a function satisfying the conditions:
(a) f (mn) = f (m)f (n) for all positive integers m and n, and
(b) f (n + 1) ≥ f (n) for all positive integers n.
Then, there is a constant α ∈ R such that f (n) = nα for all n ∈ N.
Proof We have f (1) = 1. Our job is to show that
ln f (n)
ln n
is constant when n > 1. Assume to
the contrary that
ln f (n)
ln f (m)
>
ln m
ln n
for some positive integers m, n > 1. Writing f (m) = mx and f (n) = ny , we have x > y or
ln n y
ln n
>
·
ln m
ln m x
So, we can pick a positive rational number
A
B,
(8)
(9)
where A, B ∈ N, so that
ln n
A
ln n y
>
>
· .
ln m
B
ln m x
(10)
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Hence, mA < nB and mAx > nBy . One the one hand, since f is monotone increasing, the first
A
inequality mA < nB means that f mA ≤ f nB . On the other hand, since f mA = f (m) =
B
mAx and f nB = f (n) = nBy , the second inequality mAx > nBy means that
f mA = mAx > nBy = f nB
(11)
This is a contradiction.
Fifth Solution. It is known that we get the same result when we only assume that f is monotone
increasing. In fact, in 1946, Paul Erdös proved the following result in [1]:
Theorem 1. Let f : N → R be a function satisfying the conditions:
(a) f (mn) = f (m) + f (n) for all relatively prime m and n, and
(b) f (n + 1) ≥ f (n) for all positive integers n.
Then, there exists a constant α ∈ R such that f (n) = α ln n for all n ∈ N.
This implies the following multiplicative result.
Theorem 2. Let f : N → R+ be a function satisfying the conditions:
(a) f (mn) = f (m)f (n) for all relatively prime m and n, and
(b) f (n + 1) ≥ f (n) for all positive integers n.
Then, there is a constant α ∈ R such that f (n) = nα for all n ∈ N.
Proof
1
By Proposition 5, it is enough to show that the function f is completely multi-
plicative: f (mn) = f (m)f (n) for all m and n. We split the proof in three steps.
Step 1 Let a ≥ 2 be a positive integer and let Ωa = {x ∈ N | gcd(x, a) = 1}. Then, we
obtain
f (x + a)
=1
f (x)
(12)
f ak+1 ≤ f ak f (a)
(13)
L := inf
x∈Ωa
and
for all positive integers k.
Proof of Step 1 Since f is monotone increasing, it is clear that L ≥ 1. Now, we notice that
f (k + a) ≥ Lf (k) whenever k ∈ Ωa . Let m be a positive integer. We take a sufficiently large
integer x0 > ma with gcd (x0 , a) = gcd (x0 , 2) = 1 to obtain
f (2)f (x0 ) = f (2x0 ) ≥ f (x0 + ma) ≥ Lf (x0 + (m − 1)a) ≥ · · · ≥ Lm f (x0 )
(14)
f (2) ≥ Lm .
(15)
or
Since m is arbitrary, this and L ≥ 1 force to L = 1. Whenever x ∈ Ωa , we obtain
f ak+1 f (x)
f ak+1 x
f ak+1 x + ak
=
≤
= f (ax + 1) ≤ f ax + a2 = f (a)f (x + a) (16)
f (ak )
f (ak )
f (ak )
1 We
present a slightly modified proof in [2]. For another short proof, see [3].
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or
It follows that 1 = inf x∈Ωa
f (x+a)
f (x)
f ak+1
f (x + a)
≥
.
f (x)
f (a)f (ak )
f (ak+1 )
≥ f (a)f (ak ) so that f ak+1 ≤ f ak f (a).
(17)
Step 2 Similarly, we have
f (x)
=1
f (x + a)
(18)
f ak+1 ≥ f ak f (a)
(19)
U := sup
x∈Ωa
and
for all positive integers k.
Proof of Step 2 The first result immediately follows from Step 1.
sup
x∈Ωa
f (x)
1
=
= 1.
f (x + a)
inf x∈Ωa f (x+a)
f (x)
(20)
Whenever x ∈ Ωa and x > a, we have
f ak+1 f (x)
f ak+1 x
f ak+1 x − ak
=
≥
= f (ax − 1) ≥ f ax − a2 = f (a)f (x − a). (21)
f (ak )
f (ak )
f (ak )
It therefore follows that
1 = sup
x∈Ωa
f ak+1
f (x)
f (x − a)
= sup
≤
.
f (x + a) x∈Ωa , x>a f (x)
f (a)f (ak )
Step 3 From the two previous results, whenever a ≥ 2, we have f ak+1
Then, the straightforward induction gives that
k
f ak = f (a)
(22)
= f ak f (a).
(23)
for all positive integers a and k. Since f is multiplicative, whenever
n = p1 k1 · · · pl kl
gives the standard factorization of n, we obtain
k
k
f (n) = f p1 k1 · · · f pl kl = f (p1 ) 1 · · · f (pl ) l .
(24)
(25)
We therefore conclude that f is completely multiplicative.
References
1 P. Erdos, On the distribution function of additive functions, Ann. of Math., 47(1946), 1-20
2 E. Howe, A new proof of Erdös’s theorem on monotone multiplicative functions, Amer. Math.
Monthly 93(1986), 593-595
3 L. Moser and J. Lambek, On monotone multiplicative functions, Proc. Amer. Math. Soc.,
4(1953), 544-545
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