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

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

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

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.