Question 473111
Three consecutive odd integers are such that the square of the third integer is 15 more than the sum of the squares of the first two. Find the integers.
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The numbers are x, x+2 and x+4
{{{x^2 + (x+2)^2 = (x+4)^2 - 15}}}
{{{2x^2 + 4x + 4 = x^2 + 8x + 1}}}
{{{x^2 - 4x + 3 = 0}}}
(x - 1)*(x - 3) = 0
x = 1 --> 1, 3 & 5
x = 3 --> 3, 5 & 7
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