I found the answers to be a=-1, b=0; a=1,b=0; a=0,b=1 and a=0,b=-1. Only 4 pairs, however the answer is 6. Please tell me how to proceed.
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@SamratMukhopadhyay the question is not exactly a duplicate here the RHS is 1 whereas in the prev one it is 0 – happymath May 07 '14 at 14:57
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Try the pair $(1,-1)$. – André Nicolas May 07 '14 at 14:59
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The original question has now been changed: the right-hand side that was originally 0 has now been edited to 1. – JRN May 07 '14 at 15:02
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@amWhy I've deleted my previous comment. This is the same question now as the other one (which was edited). But the wording for closing as a duplicate suggests that the duplicate question has to have an answer. And I was just noting that it didn't. I don't wholly like the wording for duplicates, and I think that repostings of essentially the same question are to be strongly discouraged. – Mark Bennet May 07 '14 at 15:12
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2If it happens again u should let the original question as it is and post another question especially when answers are already there – evil999man May 07 '14 at 15:13
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1@Mark I learned over time that if a post is an exact duplicate of a previous question with both asked by the same OP account, then the requirement that the former question have an upvoted answer is dropped. The system accepts the URL for the (non-answered) duplicated post in this scenarios (i.e. it somehow "knows" that in this situation, the questions were posted by the same user.) – amWhy May 07 '14 at 15:30
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Use quadratic formula for $a$ and you will obtain : $4-3b^2\geq 0\implies b={-1,0,1}$. For each $b$ there exist $2$ a.
evil999man
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$\left(a+\frac b2\right)^2+\frac 34b^2=1$. Now $\left(a+\frac b2\right)^2\ge0$ hence $\frac 34b^2\le1\iff4-3b^2\ge0$ or even more simply just to do as Awesome suggested. – b00n heT May 07 '14 at 15:20
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