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Question 425253: please understand this Take any number (except for 1). Square that number and then subtract one. Divide by one less than your original number. Now subtract your original number. Did you reach 1 for an answer? You should have.
Answer by tinbar(133)   (Show Source):
You can put this solution on YOUR website! you are trying to compute: ((a^2-1)/(a-1))-a for any number a in the field of Integers.
well, consider the inner bracket first; (a^2-1)/a-1: the numerator can be factored into: (a+1)(a-1) (expand and confirm this is equal to a^2-1), then our inner bracket can be rewritten as: ((a+1)(a-1))/(a-1). now there's a factor of (a-1) in the numerator and denominator, so they will cancel each other out leaving behind only (a+1). So our inner bracket is just really (a+1), then we have to subtract a from this since we had initially ignored it to handle the inner bracket, leaving behind 1.
In summary, ((a^2-1)/(a-1))-a = [{(a+1)(a-1)}/(a-1)]-a = [a+1]-a = 1 for any number a in the field of Integers
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