Question

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

Answer 1

a miracle since if you square an even integer you get an even integer and if you add two even integers together you get another even integer.

Answer 2

relax the "even" part and the numbers 4 and 5 work

Answer 3

satellite is right. Below is how you would solve the problem if you didn't specifically require even numbers:

\[a^2 + (a + 1)^2 = 41\]
\[a^2 + a^2 + 2a + 1 = 41\]
\[2a^2 + 2a + 1 = 41\]
\[2a^2 + 2a - 40 = 0\]

Since \(b^2 - 4ac = 4 - (4 \cdot 2 \cdot -40) = 4 + 320 = 324 > 0\), we know this equation has a solution.

You can use the quadratic formula to find a.

If you want to run it with even integers, you know that two even numbers must be two apart; you do the same as above:

\[a^2 + (a + 2)^2 = 41\]
\[a^2 + a^2 + 4a + 4 = 41\]
\[2a^2 + 4a + 4 = 41\]
\[2a^2 + 4a - 37 = 0\]

But, running this through the quadratic formula will leave you with a decimal number, not an integer.