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Grade: 11

                        

let t be real no. such that t^2=at+b for some positive integers a and b.then for any choice of positive integrs a and b ,t^3 is never equals to -A. 4t+3 ,B-8t+5 C 10t+3 or D 6t+5

3 years ago

Answers : (1)

Hariprasada
17 Points
							4down voteYou are close, but dont stop with t3=at2+btt3=at2+bt. Since you know what t2t2 is equal to: t3=a(at+b)+bt=(a2+b)t+abt3=a(at+b)+bt=(a2+b)t+abNow try this for every possiblility. F.ex. (A): 4=a2+b4=a2+b and ab=3ab=3 would be possible for a=1a=1 and b=3b=3.Since in the all the cases the second term is a prime number either aa or bb has to be 11 (since they are both positiv Integers) which makes it easy to check whether it`s possible or not.Edit: more detailed this gives you a quick solution if it is possible, if it doesn`t give you a quick solution we can try to get a contradiction.With the method above we see it works for (A), (C) and (D), so we want to check out (B) .Since obviously tt divides t3t3 we use the form t3=8t+5t3=8t+5 and divide it by the tt. Now we get that 55 has to be dividable by tt (note that normally this is not necessarly true, but here it holds because 8t8t is dividable by tt). Because tt has to be a positive integer it is now either 11 or 55.Case t=1t=1: Is not possible since 13≠8+513≠8+5Case t=5t=5: Is not possibe either since 53=75≠45=8∗5+553=75≠45=8∗5+5.Therefore we get a contradiction, (B) is not possible
						
2 years ago
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