In other words are there any integers that satisfy the following inequality: n² < a³ < b³ < (n +1)²? I have no idea how to solve the problem? Can someone give me a hint? I know this question is related to proof by contradiction.
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Nov 03 '19 at 09:43
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Look at $(\sqrt {a+1})^3-(\sqrt a)^3$. – David Mitra Nov 03 '19 at 09:47
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The function $f \colon \mathbb{R} \to \mathbb{R}$ defined by $$ f(x) \colon= x^3 \ \qquad \mbox{ for all } x \in \mathbb{R}$$ is a strictly increasing function with range equal to $\mathbb{R}$.
So, for $n = 0$, for example, there are two (in fact infinitely many) real numbers $a$ and $b$ cuthat $$ n^2 = 0 < a^3 < b^3 < 1 = 1^2 = (n+1)^2. $$
However, we cannot have integers $a$, $b$, and $n$ such that $$ n^2 < a^3 < b^3 < (n+1)^2. $$ In oder to show this, note that, for any integer $k$, we have $$ (k+1)^2 - k^2 = 2k+1, $$ while $$ (k+1)^3 - k^3 = 3k^2 + 3k + 1 > 2k+1.$$ Thus the cubes of integers change much more rapidly than do the squares.
Saaqib Mahmood
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