Question

Prove that there is a unique pair of consecutive odd positve integers suchi that sum of their squares is 290 find them

Answer 1

Answer:

11 ; 13

Step-by-step explanation:

a² + (a + 2)² = 290

2a² + 4a - 286 = 0

a² + 2a - 143 = 0

$a_{1}$ = - 11 ( N/A , but giving us good idea!!)

$a_{2}$ = 13

One number is 11 and another one is 13

(11)² + (13)² = 121 + 169 = 290