SOLUTION: Find three consecutive positive odd integers such that the sum of the squares of the first and second integers is equal to the square of the third integer plus 9.

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Question 1062166: Find three consecutive positive odd integers such that the sum of the squares of the first and second integers is equal to the square of the third integer plus 9.
Found 2 solutions by Alan3354, math_helper:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find three consecutive positive odd integers such that the sum of the squares of the first and second integers is equal to the square of the third integer plus 9.
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(n-2)^2 + n^2 = (n+2)^2 + 9
2n^2 - 4n + 4 = n^2 + 4n + 13
n^2 - 8n - 9 = 0
(n+1)*(n-9) = 0
n = 9
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--> 7, 9, & 11

Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!

Let x be the smaller number
x+2 = next number
x+4 = final number
+x%5E2+%2B+%28x%2B2%29%5E2+=+%28x%2B4%29%5E2+%2B+9+
+x%5E2+%2B+%28x%5E2%2B4x%2B4%29+=+%28x%5E2%2B8x%2B16%29+%2B+9+
+++2x%5E2+%2B+4x+%2B+4+=+x%5E2+%2B+8x+%2B+25+
+++x%5E2+-+4x+-+21+=+0+
+%28x-7%29%28x%2B3%29+=+0+
+x=7+ and/or +x=-3+

Check if either or both x values work:
#1 -3, -1, 1
+%28-3%29%5E2+%2B+%28-1%29%5E2+=+9%2B1++=+10+
+%281%29%5E2+%2B+9+=+10+ (ok)
#2 7, 9, 11
+7%5E2+%2B+9%5E2+=+49%2B81+=+130+
+11%5E2+%2B+9+=+121%2B9++=+130+ (ok)

Answer:
Although both answers work, I just noticed it asked for positive numbers only, so just this one answer is valid:
7, 9, 11