Question 1162614

Find three consecutive integers such that the square of the sum of the smaller two is 105 more than the square of the largest.
<pre>Let first number be N
Then 2<sup>nd</sup> and 3<sup>rd</sup> are: N + 1, and N + 2, respectively
We then get: {{{matrix(8,3, (N + N + 1)^2, "=", (N + 2)^2 + 105, (2N + 1)^2, "=", N^2 + 4N + 4 + 105, 4N^2 + 4N + 1, "=", N^2 + 4N + 109, 4N^2 - N^2 + 4N - 4N, "=", 109 - 1, 3N^2, "=", 108, N^2, "=", 108/3, N^2, "=", 36, N, "=", ""+-sqrt(36))}}}
First number or {{{highlight_green(matrix(1,3, N, "=", ""+- 6))}}}
Numbers are: {{{highlight_green(system(matrix(1,4, "6,", "7,", and, 8), OR, matrix(1,4, "- 6,", "- 5,", and, - 4)))}}}