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

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: