Phương trình, bất phương trình hàm cơ bản trên tập số tự nhiên

f : R → R R
f(x + y) = f(x) + f(y), ∀x, y ∈ R,
f (x) = ax, a ∈ R
f : R → R R
f(x + y) = f(x)f(y), ∀x, y ∈ R,
f(x) ≡ 0, f(x) ≡ 1 f (x) = a
x
, 1 = a ∈ R
+
f : R
+
→ R R
+
f(xy) = f(x) + f(y), ∀x, y ∈ R
+
,
f (x) = a ln x. a ∈ R
f : R
+
→ R R
+
f(xy) = f(x)f(y), ∀x, y ∈ R
+
,
f(x) ≡ 0, f(x) ≡ 1 f (x) = x
m

. 0 = m ∈ R
• f : X → Y X
N, N
∗
Y N, N
∗
, Z, R
•
f : N → N
f (
f (n) + f (m)
) = n + m, ∀m, n ∈ N.
f (x)
f (x) n, m ∈ N f (n) = f (m)
f (
f (n) + f (m)
) = n + m = f (
f (n) + f (n)
) = n + n→ n = m
n ∈ N, n ≥ 1
f (
f (n) + f (n)
) = 2n = (n − 1) + (n + 1) = f (
f (n −1) + f (n + 1)
)
f (n) + f (n) = f (n − 1) + f (n + 1) f
f f
f (n) = an + b
a [(an + b) + (am + b)]+b = n+ m n, m ∈ N,
a = 1, b = 0 f (n) = n

f : N → N
f (2) = 2 f (mn) = f (m) · f (n) , ∀m, n ∈ N.
f (x)
f (x) 0 ≤ f (0) < f (1) < f (2) = 2
f (0) = 0, f (1) = 1
f (3) = 3 + k, k ∈ N; f (6) = f (2) ·f (3) f (6) = 6 + 2k.
f (5) ≤ 5 + 2k f (10) = f (2) ·f (5) ≤ 10 + 4k
f (9) ≤ 9 + 4k f (18) ≤ 18 + 8k,
f (15) ≤ 15 + 8k
f (3) = 3 + k; f (5) ≥ 5 + k
f (15) = f (3) f (5) ≥ (3 + k)(5 + k).
(3 + k)(5 + k) ≤ 15 + 8k ⇔ k
2
≤ 0 ⇔ k = 0
f (3) = 3
f (2
n
+ 1) = 2
n
+ 1, n ∈ N
∗
n = 1.
f (2
n
+ 1) = 2
n
+ 1,
f

2

n+1
+ 2

= f (2) f (2
n
+ 1) = 2 (2
n
+ 1) = 2
n+1
+ 2
f

f(2
n
+ 2); f(2
n
+ 3); . . . ; f(2
n+1
+ 2)

2
n
+1
2
n
+ 1

2
n
+ 2; 2

n
+ 3; . . . ; 2
n+1
+ 2

f (2
n
+ i) = 2
n
+ i, i ∈ {2; 3; . . . ; 2
n
+ 2}
f (2
n
+ 1) = 2
n
+ 1
n + 1
n ∈ N
∗
f (n) = n n ∈ N
f (n) = n
f : N → R
f (0) = 1; f (1) = 2; f (n + 1) f
2
(n − 1) = f
3
(n) , ∀n ∈ N
∗
.

f (n)
f (n) > 0, ∀n ∈ N
ln f (0) = 0, ln f (1) = ln 2
ln f (n + 1) + 2 ln f (n − 1) = 3 ln f (n) , ∀n ∈ N
x
n
= ln f (n) , (n ∈ N)
x
0
= 0; x
1
= ln 2; x
n+2
− 3x
n+1
+ 2x
n
= 0, (n ∈ N)
λ
2
− 3λ + 2 = 0 ⇔ λ = 1 λ = 2
x
n
= A · 1
n
+ B ·2
n
= A + B · 2
n


A + B = 0
A + 2B = ln 2
⇔

A = −ln 2
B = ln 2
x
n
= (2
n
− 1) · ln 2 = ln 2
2
n
−1
f (n) .
f (n) = 2
2
n
−1
f (n) = 2
2
n
−1
(n ∈ N)
f : N → R
f (0) = 2, f (n + 1) = 3f (n) +

8f
2
(n) + 1

, ∀n ∈ N.
f (n)
f (n + 1) −3f (n) =

8f
2
(n) + 1 (≥ 1 > 0, ∀n ∈ N),
(f(n + 1) + 3f(n))
2
= 8f
2
(n) + 1.
f
2
(n + 1) + f
2
(n) = 6f (n) f (n + 1) + 1.
n n − 1
f
2
(n) + f
2
(n − 1) = 6f (n − 1) · f (n) + 1.
f
2
(n + 1) − f
2
(n − 1) = 6f (n) (f(n + 1) −f(n −1)) .
f (n) > 0 n
f(n + 1) > 3f (n) = 9f (n −1) + 3


8f
2
(n − 1) + 1 > f(n −1)
f(n + 1) − f(n − 1) > 0 f(n + 1) + f(n − 1) = 6f(n)
f(0) = 2, f(1) = 6 +
√
33, f(n + 2) −6f(n + 1) + f(n) = 0, ∀n ∈ N.
f(n) =
(8 +
√
66)(3 +
√
8)
n
8
+
(8 −
√
66)(3 −
√
8)
n
8
.
f(n)
f : N → N
f(1) > 0; f(m
2
+ n

2
) = f
2
(m) + f
2
(n), ∀m, n ∈ N.
f(n)
f(1) = f(1
2
+ 0
2
) = f
2
(1) + f
2
(0) f(0) = 0; f(1) = 1.
m = 0 f(n
2
) = f
2
(n), ∀n ∈ N
∗
f(2) = f(1
2
+ 1) = 2f
2
(1) = 2,
f(4) = f(2
2
) = f

2
(2) = 2
2
= 4,
f(5) = f(2
2
+ 1
2
) = f
2
(2) + f
2
(1) = 4 + 1 = 5,
f(25) = f(5
2
) = 25 = f(3
2
+ 4
2
) = f
2
(3) + f
2
(4) = f
2
(3) + 4,
f(3) = 3,
f(100) = f(10
2
) =


f

3
2
+ 1
2

2
=

f
2
(3) + f
2
(1)

2
=

3
2
+ 1
2

2
= 100,
f(100) = f(6
2
+ 8

2
) = f
2
(6) + f
2
(8) = f
2
(6) + f
2
(2
2
+ 2
2
)
= f
2
(6) +

f
2
(2) + f
2
(2)

2
= f
2
(6) + (4 + 4)
2
= f

2
(6) + 64
⇒ f(6) = 6.
f(n) = n, ∀n ∈ N
∗
.
n = 6.
m < n, (n ≥ 6) n = 2k + 1
(2k + 1)
2
+ (k − 2)
2
= (2k − 1)
2
+ (k + 2)
2

f((2k + 1)
2
+ (k − 2)
2
) = f
2
(2k + 1) + f
2
(k − 2)
f((2k − 1)
2
+ (k + 2)
2

) = f
2
(2k − 1) + f
2
(k + 2)
f
2
(2k + 1) + f
2
(k − 2) = f
2
(2k − 1) + f
2
(k + 2).
0 < k − 2 < k + 2 < 2k − 1 < 2k + 1 = n



f
2
(k − 2) = (k −2)
2
f
2
(2k − 1) = (2k −1)
2
f
2
(k + 2) = (k + 2)
2

f
2
(2k + 1) = (2k −1)
2
+ (k + 2)
2
− (k − 2)
2
= (2k + 1)
2
f(n) = f(2k + 1) = 2k + 1 = n
n = 2k + 2
(2k + 2)
2
+ (k − 4)
2
= (2k − 2)
2
+ (k + 4)
2
,
f(n) = f(2k + 2) = 2k + 2 = n.
f(n) = n
f(n) = n
f : N
∗
→ N
∗
f(n + 1) > f(f(n)), ∀n ∈ N
∗

f(n) = n, ∀n ∈ N
∗
.
R
f
⊆ N
∗
, R
f
= 0 R
f
f(2) > f(f(1)) > 0; f(3) > f(f(2)); . . .
R
f
f(2); f(3); . . .
f(1) R
f
f(1) ≥ 1 f(n) > 1, ∀n > 1
f : N
∗
\{1} → N
∗
\{1}.
f(2)
f(N
∗
\{1}) f(n) > 2, ∀n > 2
f : N
∗
\{1; 2} → N

∗
\{1; 2}, . . .
f(1) < f(2) < f(3) < . . . f(n) ≥ n, ∀n ∈ N
∗
f(n)
∃n ∈ N
∗
, f(n) > n f(n) ≥ n+ 1 f(f(n)) ≥ f(n+1)
f(n)
f(f(n)) < f(n + 1), ∀n ∈ N
∗
f(n) = n
f(n) = n, ∀n ∈ N
∗
.
f : N
∗
→ N
∗



f(2) = 2,
f(mn) = f(m).f(n), ∀m, n ∈ N
∗
,
f(m) < f(n), ∀m < n.
f
n = 1 f(1) = f(1.1) = f(1).f(1) f(1) = 1
2 = f(2) < f(3) < f(4) = f(2).f(2) = 4 f(3) = 3

4 = f(4) < f(5) < f(6) = f(2).f(3) = 6 f(5) = 5
f(n) = n, ∀n ∈ N
∗
f(1) = 1, f(2) = 2.
f(n) = n n = k, k ≥ 2
f(k) = k f(k + 1) = k + 1
k k + 1
f(k + 1) = f(2.
k + 1
2
) = f(2).f(
k + 1
2
) = 2.
k + 1
2
= k + 1.
k k + 2
k + 2
2
≤ k
f(
k + 2
2
) =
k + 2
2
f(k + 2) = f(2.
k + 2
2

) = f(2).f(
k + 2
2
) = 2.
k + 2
2
= k + 2.
k = f(k) < f(k + 1) < f(k + 2) = k + 2 f(k + 1) = k + 1
n = k + 1
f(n) = n, ∀n ∈ N
∗
f(n) = n
f : N
∗
→ N
∗



f(2) = 2
f(mn) = f(m).f(n), ∀m, n ∈ N
∗
f(m) < f(n), ∀m < n
; UCLN(m, n) = 1.
f
n = 1 f(1) = f(1.1) = f(1).f(1) f(1) = 1
f(3).f(5) = f(15) < f(2).f(9) < f(2).f(10) = f(2).f(2).f(5)
f(3) < f(2).f(2) = 4 2 = f(2) < f(3) < 4 f(3) = 3
f(4) = 4, f(5) = 5, f(6) = 6, f(7) = 7, f(8) = 8, f(9) = 9, f(10) = 10.
f(n) = n, ∀n ∈ N

∗
, n ≤ 10 f(n) = n, ∀n ∈ N
∗
f(k) = k (k ∈ N
∗
, 10 ≤ k ≤ n)
k = n + 1
• k = 2
α
(2l + 1), α, l ∈ N
∗
f((k + 1) + 2) = f(2
α
+ 2) = f(2(2
α−1
+ 1)) = f(2).f(2
α−1
+ 1)
• k = 2
α
, α ∈ N
∗
f(k + 2) = f(2
α
+ 2) = f(2(2
α−1
+ 1)) = f(2).f(2
α−1
+ 1) = 2(2
α−1

+ 1)
= 2
α
+ 2 = k + 2
k − 1 = f(k −1) < f(k) < f(k + 1) < f(k + 2) = k + 2
f(k) = k, f(k + 1) = k + 1.
k k + 1
• k + 1 = 2
α
(2l + 1), α, l ∈ N
∗
0 < 2
α
≤ n, 0 < 2l + 1 ≤ n.
f((k + 1) + 2) = f(2
α
+ 2) = f(2(2
α−1
+ 1)) = f(2).f(2
α−1
+ 1)
k − 1 = f(k −1) < f(k) < f(k + 1) = k + 1 f(k) = k
• k + 1 = 2
α
, α ∈ N
∗
f((k + 1) + 2) = f(2
α
+ 2) = f(2(2
α−1

+ 1)) = f(2).f(2
α−1
+ 1)
= 2(2
α−1
+ 1) = 2
α
+ 2 = (k + 1) + 2 = k + 3.
k − 1 = f(k −1) < f(k) < f(k + 1) < f(k + 2) < f(k + 3) = k + 3
f(k) = k, f(k + 1) = k + 1, f(k + 2) = k + 2
f(n) = n
f(n) = n, ∀n ∈ N
∗
f : N → N
f(1) > 0 f(m
2
+ n
2
) = f
2
(m) + f
2
(n).
f
m = n = 0 f(0) = 2f
2
(0) f(0) = 0.
n = 0 f(m
2
) = f

2
(m) f(m
2
+ n
2
) = f(m
2
) + f(n
2
)
f(1) = f
2
(1) ⇒ f(1)(1 − f(1)) = 0,
f(1) = 1 ( f(1) > 0).
f(2) = f(1
2
+ 1
2
) = f
2
(1) + f
2
(1) = 2, f(4) = f(2
2
) = f
2
(2) = 4,
f(5) = f(2
2
+ 1

2
) = 5; 25 = f(5
2
) = f(3
2
+ 4
2
) f(3) = 3.
f(6) = 6, f(7) = 7, f(8) = 8, f(9) = 9, f(10) = 10
f(n) = n n ≤ 10
f(n) = n, ∀n ∈ N
f(k) = k, k ≥ 10 f(k + 1) = k + 1
(k + 1) 5m + r, 0 ≤ r ≤ 4; m, r ∈ N
(5m)
2
= (4m)
2
+ (3m)
2
(5m + 1)
2
+ 2
2
= (4m + 2)
2
+ (3m − 1)
2
(5m + 2)
2
+ 1

2
= (4m + 1)
2
+ (3m + 2)
2
(5m + 3)
2
+ 1
2
= (4m + 3)
2
+ (3m + 1)
2
(5m + 4)
2
+ 2
2
= (4m + 2)
2
+ (3m + 4)
2
• k + 1 = 5m f
2
(5m) = f((5m)
2
) = f
2
(4m) + f
2
(3m) = (5m)

2
f(5m) = 5m.
• k+1 = 5m+1 f((5m + 1)
2
+2
2
) = f((4m + 2)
2
)+f((3m − 1)
2
)
f(5m + 1) = 5m + 1.
• k+1 = 5m+2 f((5m + 2)
2
+1
2
) = f((4m + 1)
2
)+f((3m + 2)
2
)
f(5m + 2) = 5m + 2.
• k+1 = 5m+3 f((5m + 3)
2
+1
2
) = f((4m + 3)
2
)+f((3m + 1)
2

)
f(5m + 3) = 5m + 3.
• k+1 = 5m+4 f((5m + 4)
2
+2
2
) = f((4m + 2)
2
)+f((3m + 4)
2
)
f(5m + 4) = 5m + 4.
f(k + 1) = k + 1 f(n) = n, ∀n ∈ N
f(n) = n, ∀n ∈ N
f : N
∗
→ N
∗
f(n + f(n)) = f(n), ∀n ∈ N
∗
, ∃x
0
∈ N
∗
: f(x
0
) = 1.
f
x
1

= min {x x ∈ N
∗
, f(x) = 1}.
f(x
1
+ 1) = f(x
1
+ f(x
1
)) = f(x
1
) = 1.
f(n) = 1, ∀n ∈ N
∗
, n ≥ x
1
.
x
1
> 1.
f(x
1
) − 1 + f(x
1
− 1) = f(x
1
− 1).
• x
1
− 1 + f(x

1
− 1) ≥ x
1
f(x
1
− 1) = 1
• x
1
− 1 + f(x
1
− 1) < x
1
f(x
1
− 1) < 1
f(n) = 1, ∀n ∈ N
∗
f : N
∗
→ N
∗
f(n + 1) > f(f(n)), ∀n ∈ N
∗
.
f(n) = n, ∀n ∈ N
∗
.
d f
d = min {f(n) : n ∈ N
∗

} d
m ∈ N
∗
f(m) = d
m > 1 d = f(m) > f(f(m − 1))
m = 1 f(n) m = 1
{f(n) : n ∈ N
∗
, n ≥ 2}
f(2) = min {f(n) : n ∈ N
∗
, n ≥ 2}
f(2) > f(1) f(2) = f(1) f(1) = f(2) > f(f(1))
f(1) < f(2) < f(3) < f(4) < ··· < f(n) < . . .
f(n) ∈ N
∗
f(1) ≥ 1 f(k) ≥ k
f(k) > k f(k) ≥ k + 1
f(k + 1) > f(f(k)).
f(k) ≥ k + 1 f(k + 1) ≤ f(f(k))
f(k) > k
f(k) = k, ∀k ∈ N
∗
f(n) ≡ n
f(n) = n, ∀n ∈ N
∗
f : N
∗
→ N
∗

f(1) = 1 f(f(n))f(n + 2) + 1 = f(n + 1)f(f(n + 1)), ∀n ∈ N
∗
.
f
n ∈ N
∗
f(n + 1) > f(f(n)).
n = 1
n = k, (k ≥ 1)
n = k + 1.
f(f(k))f(k + 2) = f(k + 1)f(f(k + 1)) − 1.
f(k + 1) =
f(k + 1)f(f(k + 1)) −1
f(f(k))
≥
(f(f(k)) + 1)f(f(k + 1)) − 1
f(f(k))
>
f(f(k))f(f(k + 1))
f(f(k))
= f(f(k + 1)).
f(n + 1) > f(f(n)), ∀n ∈ N
∗
f(n) = n, ∀n ∈ N
∗
.
f(n) = n, ∀n ∈ N
∗
f ∈ C(R)
f(

x + y
2
) =
f(x) + f(y)
2
, ∀x, y ∈ R.
• f : X → Y X
N, N
∗
Y N, N
∗
, Z, R
•
f : N → Z
f(m + f
2
(m + 1)) = −f
2
(m + 1) − (m + 1), ∀m ∈ N.
f(n)
k = f
2
(m + 1) k ∈ N f(m + 1) ∈ Z
f(m + k) = −k −m −1 = −(m + k) −1 f(n) = −n − 1
f(n) = −n − 1(n ∈ N)
f : N → N
f(f(n)) + f(n) = 2n + 3, ∀n ∈ N.
f(n)
n = 0 f(f(0)) + f(0) = 3 f(0) ≤ 3.
f(0) = 0 0 = 3

f(0) = 2 f(f(0)) = 1 ⇒ f(2) = 1
f(1) = f(f(2)) = 2.2 + 3 − f(2) = 6.
n = 1 f(f(1)) = 2.1 + 3 − f(1) = −1 f(6) = −1.
f(n) ∈ N.
f(0) = 3 f(3) = f(f(0)) = 2.0 + 3 −f(0) = 0 f(3) = 0.
2.3 + 3 = f(f(3)) + f(3) = f(0) + 0 = 3 9 = 3
f(0) = 1
f(n) = n + 1, ∀n ∈ N.
n = 0.
n = k, (k ≥ 0) f(k) = k + 1
f(k + 1) = f(f(k)) = 2k + 3 − f(k) = 2k + 3 − (k + 1) = k + 2
f(k + 1) = k + 1 + 1.
n = k + 1
n ∈ N
f(n) = n +1, ∀n ∈ N
f(n) = n + 1, ∀n ∈ N
f : N → N
f(f(n)) + f(n) = 2n + 3k, ∀n ∈ N,
k
f
a
1
= n n ≥ 1 a
n+1
= f(a
n
)
2a
n
+ 3k = a

n+1
+ a
n+2
,
2a
n+1
+ 3k = a
n+2
+ a
n+3
.
a
n+3
− 3a
n+1
+ 2a
n
= 0
a
n
= λ
1
+ nλ
2
+ λ
3
(−2)
n
, ∀n ∈ N
∗

.
λ
3
> 0 n a
n
< 0
λ
3
< 0 n a
n
< 0
λ
3
= 0 a
n
= λ
1
+ nλ
2
2λ
1
+ 2nλ
2
+ 3k = λ
1
+ (n + 1)λ
2
+ λ
1
+ (n + 2)λ

2
λ
2
= k
a
2
− a
1
= λ
1
+ 2k − (λ
1
+ k) = k f(n) − n = k
f(n) = n + k, ∀n ∈ N
f(n) = n + k, ∀n ∈ N
f : N → N
f(f(f(n))) + 6f(n) = 3f(f(n)) + 4n + 2007, ∀n ∈ N.
f k
(x
n
) x
0
= k x
n+1
= f(x
n
)
n x
n
x

n+3
= 3x
n+2
− 6x
n+1
+ 4x
n
+ 2007, ∀n ∈ N.
λ
3
− 3λ
2
+ 6λ − 4 = 0 ⇔ λ ∈

1, 1 ±i
√
3

(x
n
)
x
n
= A + 2
n

B cos
nπ
3
+ C sin

nπ
3

+ Dn, ∀n ∈ N
k = x
0
= A + B
f(k) = f(x
0
) = x
1
= A + B + C
√
3 + D ⇒ f(k) = k + C
√
3 + D.
D + C
√
3 = 669
f(n) = n + 669, ∀n ∈ N.
f : N
∗
→ N
∗
f(n + f(n)) = 2f(n), ∀n ∈ N
∗
.
f
f f(n + 1) ≥ f(n) + 1
f(n + 1) − n − 1 ≥ f(n) − n, ∀n ∈ N

∗
f(n) − n
a
0
= 1, a
n+1
= a
n
+ f(a
n
)
a
0
< a
1
< . . . , f(a
n+1
) = 2f(a
n
)
f(a
n+1
) − a
n+1
= f(a
n
) − a
n
(m, n) f(n)−n = f(m)−m
, f(n) − n

f(n) − n N
∗
f(n) = n + k, ∀n ∈ N
∗
k
f : N
∗
→ N
∗
f(2x + 3y) = 2f(x) + 3f(y) + 4, ∀x, y ∈ N
∗
.
P (u, v) x u y v
p(x + 3, y)
f(2(x + 3) + 3y) = 2f(x + 3) + 3f(y) + 4, ∀x, y ∈ N
∗
.
p(x, y + 2)
f(2x + 3(y + 2)) = 2f(x) + 3f(y + 2) + 4, ∀x, y ∈ N
∗
.
2 [f(x + 3) −f(x)] = 3 [f(y + 2) − f(y)] , ∀x, y ∈ N
∗
.
y = 1
2 [f(x + 3) −f(x)] = 3 [f(3) −f(1)] , ∀x ∈ N
∗
.
2 3 f(x + 3) − f(x)
f(x + 3) − f(x) = 3c, ∀x ∈ N

∗
c
3 [f(y + 2) − f(y)] = 6c, ∀y ∈ N
∗
f(y + 2) − f(y) = 2c, ∀y ∈ N
∗

f(x + 2) = f(x) + 2c, ∀x ∈ N
∗
f(x + 3) − f(x) + 3c, ∀x ∈ N
∗

f(x + 3) = f(x + 1) + 2c, ∀x ∈ N
∗
f(x + 3) = f(x) + 3c, ∀x ∈ N
∗
f(x + 1) = f(x) + c, ∀x ∈ N
∗
{f(x)}
+∞
x=1
f(x) = cx + d, ∀x ∈ N
∗
c(2x + 3y) + d = 2(cx + d) + 3(cy + d) + 4, ∀x ∈ N
∗
d = −1
f(x) = cx−1, ∀x ∈ N
∗
c > 1
f : N → N f(0) = 1

f(f(n)) + 3f(n) = 4n + 5, ∀n ∈ N.
n ∈ N
f(n) = n + 1.
f(0) = 1 = 0 + 1 n = 0.
n = k, (k ∈ N) f(k) = k +1.
n = k + 1 f(k + 1) = k + 2.
f(k+1) = f(f(k))
do
=
4k+5−3f(k) = 4k +5−3(k+1) = k+2.
f(n) = n + 1, ∀n ∈ N
f : N
∗
→ N
∗
f(mn) + f(m + n) = f(m)f(n) + 1, ∀m, n ∈ N
∗
.
x = f(1), y = f(2), z = f(3), t = f(4)
p(u, v) m u n v
p(1, 1) x + y = x
2
+ 1.
p(1, 2) y + z = xy + 1.
p(1, 3) z + t = xz + 1.
p(2, 2) 2t = y
2
+ 1.
2z + y
2

+ 1 = 2xz + 2 ⇔ z =
y
2
− 1
2(x − 1)
x
2
− x + 1 +
(x
2
− x + 1)
2
− 1
2(x − 1)
= x(x
2
− x + 1) + 1
⇔ x
2
− x +
(x
2
− x + 1)
2
− 1
2(x − 1)
= x
3
− x
2

+ x
⇒ 2x
3
− 4x
2
+ 2x + (x
2
− x + 1)
2
− 1 = 2(x
4
− 2x
3
+ 2x
2
− x)
⇒ 6x
3
− 8x
2
+ 4x + (x
2
− x + 1)
2
− 1 = 2x
4
⇒ 6x
3
− 8x
2

+ 4x + x
4
+ x
2
− 2x
3
− 2x + 2x
2
= 2x
4
⇒ x
4
− 4x
3
+ 5x
2
− 2x = 0
⇒ x(x −1)
2
(−2) = 0 ⇒ x ∈ {1, 2}, (x = f(1) ∈ N
∗
).
p(1, n), f(n) + f(n + 1) = f(1)f(n) + 1, ∀n ∈ N
∗
f(n + 1) = [f(1) − 1] f(n) + 1, ∀n ∈ N
∗
.
f(1) = 1 f(n) = 1, ∀n ∈ N
∗
f(1) = 2

f(n+1) = f(n)+1, ∀n ∈ N
∗
f(n) = f(1)+(n−1) = n+1, ∀n ∈ N
∗
f(n) = 1, ∀n ∈ N
∗
; f(n) = n + 1, ∀n ∈ N
∗
f ∈ C(R)
f(x + y) + f(x − y) = 2f(x)f(y), ∀x, y ∈ R
• f(0) = 0 f ≡ 0.
• f(0) = 1 f ≡ 1, |f(x)| ≤ 1; ∀x ∈ R f(x) = cos ax.
• f(0) = 1, ∃c ∈ R : |f(c)| > 1 f(x) = ±cosh x.
• f : X → Y X
N, N
∗
Y N, N
∗
, Z, R
•
f : N
∗
→ N
∗
n
0
∈ N
∗
f(n
0

) = 1 f(n + f(n)) = f(n), ∀n ∈ N
∗
.
f(n)
n f(n) = 1 f(n+ 1) = f(n +f(n))
(1.29)
=
f(n) = 1
n
0
∈ N
∗
f(n
0
) = 1
f(n) = 1 n ∈ N
∗
S := {n ∈ N
∗
f(n) = 1} S
S = ∅ f(n) = 1, ∀n ∈ N
∗
S = ∅ S n
1
S
f(n
1
+ f(n
1
)) = f(n

1
) = 1 n
1
+ f(n
1
) ∈ S
f(n
1
) ∈ N
∗
n
1
+ f(n
1
) > n
1
n
1
+ f(n
1
) /∈ S
n
1
S
S = ∅ f ≡ 1, (n ∈ N
∗
)
f : N → R
f(0) = β; f(n + 1) = 2f
2

(n) − 1, ∀n ∈ N.
f(n)
β = 1
f(0) = 1; f(1) = 2f
2
(0) − 1 = 1; f(2) = 2f
2
(1) − 1; f(3) = 2f
2
(2) − 1.
∀n ∈ N, f(n) = 1
• β = −1 ∀n ∈ N, f(n) = 1
• |β| < 1 ϕ cos ϕ = β ⇔ ϕ = cos β
f(0) = cos ϕ = cos 2
0
ϕ
f(1) = 2f
2
(0) − 1 = 2cos
2
ϕ − 1 = cos 2
1
ϕ
f(2) = 2f
2
(1) − 1 = 2cos
2
2ϕ − 1 = cos 2
2
ϕ

f(3) = 2f
2
(2) − 1 = 2cos
2
2
2
ϕ − 1 = cos 2
3
ϕ.
∀n ∈ N, f(n) = cos 2
n
ϕ
|β| > 1 ϕ cosh ϕ = β
f(0) = c ϕ = c 2
0
ϕ
f(1) = 2f
2
(0) − 1 = 2cosh
2
ϕ − 1 = c 2
1
ϕ
f(2) = 2f
2
(1) − 1 = 2cosh
2
2ϕ − 1 = c 2
2
ϕ

f(3) = 2f
2
(2) − 1 = 2cosh
2
2
2
ϕ − 1 = c 2
3
ϕ
∀n ∈ N, f(n) = c 2
n
ϕ
cosh ϕ = β
e
ϕ
+ e
−ϕ
2
= β
e
ϕ
= β −

β
2
− 1 e
ϕ
= β +

β

2
− 1.
f(n) =
1
2

(e
ϕ
)
2
n
+
1
(e
ϕ
)
2
n

=
1
2

(β −

β
2
− 1)
2
n

+ (β +

β
2
− 1)
2
n

f(n) =













1, (β = 1)

−1(n = 0)
1(n ≥ 1)
β = −1
cos 2
n
β, (|β| < 1)

±1
2

(β −

β
2
− 1)
2
n
+ (β +

β
2
− 1)
2
n

, (|β| > 1)
f : N → R
f(n)f(m) = f(m + n) + f(n − m), ∀n, m ∈ N, n ≥ m.
f(n)
m = n = 0 f
2
(0) = 2f(0) f(0) = 0 ∨ f(0) = 2
• f(0) = 0 m = 0 2f(n) = 0 f(n) = 0, ∀n ∈ N
• f(0) = 2 m = 1 a := f(1), x
n
:= f(n)
x

0
= 2; x
1
= a; x
n+2
− ax
n+1
+ x
n
= 0, (n ≥ 1).
t
2
− at + 1 = 0,
∆ = a
2
− 4.
• ∆ > 0 x
n
= λ
n
+
1
λ
n
λ = a +
√
a
2
− 4 ∈ R
∗

• ∆ = 0 x
n
= 2(−1)
n
= 2 cos nπ
• ∆ < 0 x
n
= 2 cos nϕ, (ϕ ∈ R).
f ≡ 0; f(n) = λ
n
+
1
λ
n
, (λ ∈ R
∗
); f(n) = 2 cos nϕ, (ϕ ∈ R).
f : N
∗
→ N
∗
2(f(m
2
+ n
2
))
3
= f
2
(m)f(n) + f(m)f

2
(n), ∀m, n ∈ N
∗
.
f
f(n) ≡ c c
m, n ∈ N
∗
f(m) = f(n) a, b
|f(a) − f(b)| = min |f(m) −f(n)|, m, n ∈ N
∗
.
f(a) > f(b) 2f
3
(b) < f
2
(a).f(b) + f(a).f
2
(b) < 2f
3
(a)
f(b) < f(a
2
+ b
2
) < f(a) f(a
2
+ b
2
) − f(b) < f(a) − f(b)

|f(a) − f(b)| = f(a) − f(b) > f(a
2
+ b
2
) − f(b) =


f(a
2
+ b
2
) − f(b)


.
f(n) ≡ c c
f : N
∗
→ N
∗
x.f(y) + yf(x) = [xf(f(x)) + yf(f(y))] f(xy), ∀x, y ∈ N
∗
.
x = y = 1
2f(1) = 2f(1)f(f(1))
dof(1)≥1
⇔
f(f(1)) = 1.
f 1 ≤ f(1)
1 ≤ f(1) ≤ f(f(1)) = 1,

f(1) = 1.
y = 1
x + f(x) = [xf(f(x)) + 1] f(x), ∀x ∈ N
∗
⇔ x = xf(f(x))f(x), ∀x ∈ N
∗
⇔ f(f(x))f(x) = 1, ∀x ∈ N
∗
⇒ f(x) = f(f(x)) = 1, ∀x ∈ N
∗
f : N
∗
→ N
∗
f(x) = 1, ∀x ∈ N
∗
.