Question 1031151
Original Problem:
Choose 3 consecutive integers. Square the middle integer, then multiply the other 2 integers. Subtract the product from the squared number. The answer is 1. Why does this work? Is there an equation for this?
2, 3, 4
9-8 = 1
Thank you
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Explanation:
<font color=blue>Choose 3 consecutive integers</font> tells us that we have 3 numbers like 1,2,3 or 4,5,6 or 7,8,9. It could be any 3 numbers where one follows right after the other. In general, this means
x = first number
y = second number
z = third number


Since y follows right after x, we know y = x+1. Similarly, z follows right after y, so
z = y+1
z = (x+1)+1 ... replace y with x+1
z = x+2


So far we have
y = x+1
z = x+2


<font color=blue>Square the middle integer</font> so let's square y to get
y = x+1
y^2 = (x+1)^2 = x^2+2x+1


<font color=blue>then multiply the other 2 integers</font> so we'd get x*z = x*(x+2) = x^2+2x


<font color=blue>Subtract the product from the squared number</font>. Now they want us to do this subtraction
(squared value) - (product) = (x^2+2x+1) - (x^2+2x) = x^2+2x+1-x^2-2x = 1


So that is why the answer is always 1. I see you've done it with fixed numbers, which is great, but it's best to do it algebraically to show that it works for ANY 3 consecutive integers. 


Let me know if you have any questions.