Question

The sum of the four consecutive squares 21^2+ 22^2+ 23^2+24^2 is equal to the sum of three consecutive integers squared. What are they?

Answer 1

The sum of squares of four consecutive numbers 21, 22, 23, and 24 is equal to the sum of squares of three consecutive integers 13, 14, and 15.

Step-by-step explanation:

The question asks to find three consecutive integers whose squares sum up to the sum of the squares of four consecutive numbers, specifically 212, 222, 232, and 242. Let's calculate this sum:

• 212 = 441
• 222 = 484
• 232 = 529
• 242 = 576

Adding them up gives us 2030.

Now, let us denote the three consecutive integers as n, n+1, and n+2. The sum of their squares is n2 + (n+1)2 + (n+2)2. This can be written as:

n2 + n2 + 2n + 1 + n2 + 4n + 4 = 2030

Combining like terms:

3n2 + 6n + 5 = 2030

Subtracting 2030 from both sides:

3n2 + 6n - 2025 = 0

Solving this quadratic equation, we find that n = 13.

Thus, the three consecutive integers are 13, 14, and 15, as 132 + 142 + 152 = 169 + 196 + 225 = 590, which indeed equals 2030.