SOLUTION: find 3 consecutive integers such that the square of the third exceeds the sum of the secod and the square of the first by 42

Question 157022: find 3 consecutive integers such that the square of the third exceeds the sum of the secod and the square of the first by 42
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
Consecutive numbers are: x,(x+1),(x+2)
(x+2)^2=(x+1)+x^2+42
x^2+4x+4=x+1+x^2+42
x^2-x^2+4x-x=42+1-4
3x=39
x=39/3
x=13 for the smaller integer.
13+1=14 for the second.
13+2=15 for the third.
Proof:
15^2=14+13^2+42
225=14+169+42
225=225

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