SOLUTION: Find three consecutive nunbers where the product of two smaller numbers is 37 less than the square of the largest number

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Question 40962: Find three consecutive nunbers where the product of two smaller numbers is 37 less than the square of the largest number
Found 2 solutions by psbhowmick, josmiceli:
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
Let the numbers be (n-1), n, (n+1).
Smaller two numbers: n and (n-1); their product = n(n-1)
Largest number: (n+1); its square %28n%2B1%29%5E2
So %28n%2B1%29%5E2+-+n%28n-1%29+=+37
or n%5E2+%2B+2n+%2B+1+-+n%5E2+%2B+n+=+37
or 3n = 37 - 1 = 36
or n+=+36%2F3 = 12
Hence the numbers are (12-1=) 11, 12 & (12+1=) 13.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The numbers are n, n+1, n+2
n(n+1) = (n+2)^2 - 37
n^2 + n = n^2 +4n + 4 -37
n = 4n -33
-3n = -33
3n = 33
n = 11
The numbers are 11,12,13
check
n(n+1) = (n+2)^2 - 37
11 * 12 = 13^2 -37
132 = 169 - 37
132 = 132
OK