SOLUTION: Find three consecutive odd integers such that the sum of the squares of the first and third is 89 more than the square of the second.

Question 901187: Find three consecutive odd integers such that the sum of the squares of the first and third is 89 more than the square of the second.
Answer by JulietG(1812) (Show Source): You can put this solution on YOUR website!
There are many ways to do this. I did it the easiest way for me.
Take the square root of 89. It's closest to 9. That's your middle number.
7,9,11 is the sequence.

Prove it:the sum of the squares of the first and third is 89 more than the square of the second.
+=+89
49 + 121 = 81 + 89
170 = 170

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