SOLUTION: Find three consecutive odd integers such that the product of the first and third is equal to 1 less than twice the second. It's quite a confusing problem to read, but I wasn't

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Question 1009159: Find three consecutive odd integers such that the product of the first and third is equal to 1 less than twice the second.
It's quite a confusing problem to read, but I wasn't taught this in class because I recently moved schools.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!

n is any positive integer.
2n is a positive even integer.
2n+1 is an ODD positive integer.

The description transcribed into an equation:
highlight_green%28%282n%2B1%29%282n%2B5%29=-1%2B2%282n%2B3%29%29

Solve for n;
evaluate system%282n%2B1%2C2n%2B3%2C2n%2B5%29.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Find three consecutive odd integers such that the product of the first and third is equal to 1 less than twice the second.
It's quite a confusing problem to read, but I wasn't taught this in class because I recently moved schools.
Let the smallest be S

Then others are: S + 2, and S + 4

We then get: S(S + 4) = 2(S + 2) - 1

S%5E2+%2B+4S+=+2S+%2B+4+-+1

S%5E2+%2B+4S+-+2S+-+4+%2B+1+=+0

S%5E2+%2B+2S+-+3+=+0

(S + 3)(S - 1) = 0                 

S, or smallest = highlight_green%28-+3%29        OR         highlight_green%28S+=+1%29

You should be able to find the other 2