let [root(n 2 +1)]=[root(n 2 +λ)] , where n, λ€N . show that λ can have 2n different values.

Question

let [root(n 2 +1)]=[root(n 2 +&lambda;)] , where n, &lambda;&euro;N . show that &lambda; can have 2n different values.

mycroft holmes · Answer

for 1&le; &lambda;&le; 2n we have n 2 < n 2 +&lambda;<(n+1) 2 Hence n < &radic;(n 2 +&lambda;) < (n+1) and hence we have [&radic;(n 2 +&lambda;)] = [&radic;(n 2 +1)] = n If &lambda;&le;0 or &lambda;>2n, [&radic;(n 2 +&lambda;)] &ne; n So these are the 2n values of &lambda; for which the given equation holds

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let [root(n 2 +1)]=[root(n 2 +λ)] , where n, λ€N . show that λ can have 2n different values.

let [root(n²+1)]=[root(n²+λ)] , where n, λ€N . show that λ can have 2n different values.

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let [root(n 2 +1)]=[root(n 2 +λ)] , where n, λ€N . show that λ can have 2n different values.

let [root(n2+1)]=[root(n2+λ)] , where n, λ€N . show that λ can have 2n different values.

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let [root(n²+1)]=[root(n²+λ)] , where n, λ€N . show that λ can have 2n different values.