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
<h3>Answer: 4 and 5</h3>
Other method:
n, n + 1 - two consecutive positive integers
The equation:
n² + (n + 1)² = 41 <em>use (a + b)² = a² + 2ab + b²</em>
n² + n² + 2(n)(1) + 1² = 41
2n² + 2n + 1 = 41 <em>subtract 41 from both sides</em>
2n² + 2n - 40 = 0 <em>divide both sides by 2</em>
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
<h3>Answer: 4 and 5.</h3>
