The number of functions f: 1, 2, 3, 4 → a ∈ ℤ: | a | ≤ 8 satisfying...

Given,

$f ( n ) + \dfrac{1}{n}f ( n + 1 ) = 1, \forall n \in { 1, 2, 3 }$

$⇒ nf ( n ) + f ( n + 1 ) = n$

At $n = 1$,

$f ( 1 ) + f ( 2 ) = 1$ ....(1)

At $n = 2$,

$2f ( 2 ) + f ( 3 ) = 2$ ....(2)

At $n = 3$,

$3f ( 3 ) + f ( 4 ) = 3$ ....(3)

Put the value of $f ( 2 )$ from equation (1) in equation (2),

$2 ( 1 - f ( 1 ) ) + f ( 3 ) = 2$

$⇒ f ( 3 ) = 2f ( 1 )$ ....(4)

Put the value of $f ( 3 )$ from equation (4) in equation (3),

$3 ( 2f ( 1 ) ) + f ( 4 ) = 3$

$⇒ f ( 4 ) = 3 - 6f ( 1 )$

$∵ f: { 1, 2, 3, 4 } \to { a \in \mathbb{Z}: | a | \le 8 }$

$⇒ - 8 \le f ( 4 ) \le 8$

$⇒ - 8 \le 3 - 6f ( 1 ) \le 8$

$⇒ - 11 \le - 6f ( 1 ) \le 5$

$⇒ \dfrac{- 5}{6} \le f ( 1 ) \le \dfrac{11}{6}$

$⇒ f ( 1 ) = 0, 1$

Case I: $f ( 1 ) = 0$

$⇒ f ( 2 ) = 1, f ( 3 ) = 0, f ( 4 ) = 3$

Case II: $f ( 1 ) = 1$

$⇒ f ( 2 ) = 0, f ( 3 ) = 2, f ( 4 ) = - 3$

Therefore, two such functions are possible.