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PEN Math problems from the collection "Problems in Elementary Number Theory"

Math problems from the collection "Problems in Elementary Number Theory"

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PEN Math problems from the collection "Problems in Elementary Number Theory"

Math problems from the collection "Problems in Elementary Number Theory"

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Peter

#1 h May 25, 2007, 12:25 AM • 2 Y

Y by Adventure10, Mango247

Do there exist irrational numbers $a, b>1$ and $\lfloor a^{m}\rfloor \not=\lfloor b^{n}\rfloor$ for any positive integers $m$ and $n$ ?

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Peter

#2 h Aug 30, 2007, 12:15 PM • 1 Y

Y by Adventure10

Yes, take for example $a =\sqrt{3}$ and $b =\sqrt{6}$ and assume that $\lfloor a^{m}\rfloor =\lfloor b^{n}\rfloor = k$ for some positive integers $m,n,k$ . Then we need $k^{2}\le 3^{m},6^{n}< (k+1)^{2}$ , so that $|3^{m}-6^{n}|\le 2k$ . Since obviously $m > n$ we have that $2k\ge 3^{n}.|3^{m-n}-2^{n}|\ge 3^{n}$ . But then $4.6^{n}\le 4k^{2}\le 9^{n}$ , so $4\le\left(\frac{3}{2}\right)^{n}$ , so $n\le 3$ . But $(\lfloor b^{n}\rfloor) = 2,6,14,...$ , while $(\lfloor a^{m}\rfloor) = 1,3,5,9,15,...$ , so no terms are equal, QED.

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hxy09

#3 h Feb 21, 2009, 2:41 PM • 2 Y

Y by Adventure10, Mango247

Peter wrote:

Yes, take for example $a = \sqrt {3}$ and $b = \sqrt {6}$ and assume that $\lfloor a^{m}\rfloor = \lfloor b^{n}\rfloor = k$ for some positive integers $m,n,k$ . Then we need $k^{2}\le 3^{m},6^{n} < (k + 1)^{2}$ , so that $|3^{m} - 6^{n}|\le 2k$ . Since obviously $m > n$ we have that $2k\ge 3^{n}.|3^{m - n} - 2^{n}|\ge 3^{n}$ . But then $4.6^{n}\le 4k^{2}\le 9^{n}$ , so $4\le\left(\frac {3}{2}\right)^{n}$ , so $n\le 3$ . But $(\lfloor b^{n}\rfloor) = 2,6,14,...$ , while $(\lfloor a^{m}\rfloor) = 1,3,5,9,15,...$ , so no terms are equal, QED.

Sorry I don't quite catch for the statement $k^{2}\le 3^{m},6^{n} < (k + 1)^{2}$ , so that $|3^{m} - 6^{n}|\le 2k$
But we can choose $a = 2 + \sqrt {3}$ $b = \frac {5 + \sqrt17}{2}$ easily prove $\lfloor a^{m}\rfloor$ is odd and $\lfloor b^{n}\rfloor$ is even.Leads to contradiction.

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