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Find the number of positive integers ${}n$ that satisfy $\lfloor \sqrt{n} \rfloor + \lfloor \sqrt{4n} \rfloor = 5.$

 Dec 10, 2023
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Since ⌊n​⌋≤n​<⌊n​⌋+1, we have $⌊n​⌋2≤n<(⌊n​⌋+1)2.$Since ⌊4n​⌋≤4n​<⌊4n​⌋+1, we have $4⌊n​⌋2≤4n<4(⌊n​⌋+1)2.$

 

Adding these inequalities, we get $5⌊n​⌋2<5n<5(⌊n​⌋+1)2.$

 

Dividing all terms by 5, we get $⌊n​⌋2+51​ <(⌊n​⌋+1)2+1.$  Since ⌊n​⌋+⌊4n​⌋=5, $(⌊n​⌋+1)2+1=⌊n​⌋2+2⌊n​⌋+2=5−(⌊n​⌋−1)2<5.$

 

Hence, we have $⌊n​⌋2+51​ <5.$  Since n is an integer, we can write this as $⌊n​⌋2+1≤n≤4.$

 

We can check (by squaring the numbers from 1 to 4) that n=2​ and n=4​ are the only integers that satisfy this inequality.

 Dec 11, 2023

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