The sum of the squares of two consecutive natural numbers is 41. Find the numbers.

Sure, let's solve this step by step.

Step 1: Let's denote the two consecutive natural numbers as n and n+1.

Step 2: According to the problem, the sum of the squares of these two numbers is 41. So, we can write the equation as:

n^2 + (n+1)^2 = 41

Step 3: Expand the equation:

n^2 + n^2 + 2n + 1 = 41

Step 4: Simplify the equation:

2n^2 + 2n + 1 - 41 = 0

Step 5: Further simplify the equation:

2n^2 + 2n - 40 = 0

Step 6: Divide the equation by 2 to make it simpler:

n^2 + n - 20 = 0

Step 7: This is a quadratic equation in the form of ax^2 + bx + c = 0. We can solve it by factoring:

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

Step 8: Set each factor equal to zero and solve for n:

n - 4 = 0 => n = 4

n + 5 = 0 => n = -5

Step 9: Since n is a natural number, n = -5 is not a valid solution. So, the two consecutive numbers are 4 and 5.