SOLUTION: The sum of the squares of two consecutive negative integers is 100. Find the integers. (I got the answer but i don't know how to solve it, I did guess and check)

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Question 140735: The sum of the squares of two consecutive negative integers is 100. Find the integers. (I got the answer but i don't know how to solve it, I did guess and check)
Found 2 solutions by checkley77, solver91311:
Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
thnks but I got 8 and 13
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Can't be the correct answers because they are not consecutive integers and they don't work in the proof:
8^2+13^2=64+169=233 not 100.
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(-x)^2+(-x+1)^2=100
x^2+x^2-2x+1=100
2x^2-2x+1-100=0
x^2-2x-99=0
using the quadratic equation we get:
x=(2+-sqrt[-2^2-4*2*-99])/2*2
x=(2+-sqrt[4+792)/4
x=(2+-sqrt796)/4
x=(2+-28.2)/4
x=(2-28.2)/4
x=-26.2/4
x=-6.55 answer.
-6.55+1=-5.55 answer.

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!
Um, sorry to burst your bubble, but I don't think you got the answer required. That's because the answer does not exist.

Trial and error proof:
-6 * -6 = 36
-7 * -7 = 49
36 + 49 = 85 < 100
-7 * -7 = 49
-8 * -8 = 64
49 + 63 = 113 > 100

Algebraic proof:

First integer: x
Next integer: x + 1

Sum of the integer's squares is, meaning equals, 100:

but is irrational, therefore x cannot be an integer.

Now had the problem been 'two consecutive negative EVEN integers' that would be another story.