Question

The sum of the squares of two consecutive positive integers is 41. Find the two

Answer 1

We have to present the number 41 as the sum of two squares of consecutive positive integers.

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25

16 + 25 = 41

Answer: 4 and 5

Other method:

n, n + 1 - two consecutive positive integers

The equation:

n² + (n + 1)² = 41     use (a + b)² = a² + 2ab + b²

n² + n² + 2(n)(1) + 1² = 41

2n² + 2n + 1 = 41     subtract 41 from both sides

2n² + 2n - 40 = 0     divide both sides by 2

n² + n - 20 = 0

n² + 5n - 4n - 20= 0

n(n + 5) - 4(n + 5) = 0

(n + 5)(n - 4) = 0 ↔ n + 5 = 0 ∨ n - 4 =0

n = -5 < 0 ∨ n = 4 >0

n = 4

n + 1 = 4 + 1 = 5

Answer: 4 and 5.