Question

Determinați toate numerele naturale a b care au proprietatea ca 9 supra a²+b² reprezinta un numar natural.​

Answer 1

Răspuns:

(a,b) ∈ {(0,1);(0,3);(1,0);(3,0)}

Explicație pas cu pas:

$\frac{9}{ {a}^{2} + {b}^{2}} \in \mathbb{N} , \ a \in \mathbb{N} , \ b \in \mathbb{N} \\ \implies ({a}^{2} + {b}^{2})\in\{1;3;9\} \\0 \leqslant {a}^{2} \leqslant 9\implies 0 \leqslant a \leqslant 3 \\ 0 \leqslant {b}^{2} \leqslant 9\implies 0 \leqslant b \leqslant 3$

1)

a = 0 => a² = 0² = 0

b = 1 => b² = 1² = 1

a² + b² = 0 + 1 = 1

2)

a = 0 => a² = 0² = 0

b = 3 => b² = 3² = 9

a² + b² = 0 + 9 = 9

3)

a = 1 => a² = 1² = 1

b = 0 => b² = 0² = 0

a² + b² = 1 + 0 = 1

4)

a = 3 => a² = 3² = 9

b = 0 => b² = 0² = 0

a² + b² = 9 + 0 = 9

=>

(a,b) ∈ {(0,1);(0,3);(1,0);(3,0)}