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In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that-

$(xz)^k\leq b^3 \implies (a+1)(ab^2+1) \leq b^3$ , I could not figure it out. Here, $(xz)^k=(a+1)(ab^2+1)$ (see page $8$),$a, b \geq 2, $ and both are integers, $k\geq 3 $ (see theorem 11 on page $8$, please see the attached-paper for detail ).

If we simplify by Rearrangement inequality, we have that

$$(a+1)(ab^2+1)=a^2b^2+ab^2+a+1 $$

how this could be less than or equal to $b^3$?

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  • $\begingroup$ You should clarify better the conditions for a and b otherwise we need to study the full paper to answer properly. $\endgroup$ – gimusi Sep 8 '18 at 9:15
  • $\begingroup$ @gimusi $a, b \geq 2, $ , both are integers (see theorem 11 on page $8$). $\endgroup$ – Mike SQ Sep 8 '18 at 9:19
  • $\begingroup$ is that the only condition? For $a=b=2$ the inequality clearly doesn't hold therefore I suppose the must be more conditions on a and b. $\endgroup$ – gimusi Sep 8 '18 at 9:23
  • $\begingroup$ @gimusi there is an assumption( 2 lines above the statement) , that $a \geq 2^{49}$, but I dont see how that is related , it would be better if u see the paper yourself $\endgroup$ – Mike SQ Sep 8 '18 at 9:28
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On page 8 there is the inequality

$$ b\geq (k^{k}a^{k-2})^{1/2} \label{12}\tag{12} $$

and due to the comment on page 9 we may assume that $k\geq 50$ and $a\geq 2^{49}$. Since $b\geq 1$, we see by \eqref{12} that

$$b^{3}\geq b^{2}(k^{k}a^{k-2})^{1/2} \label{*}\tag{*}$$

Now we have, since $b\geq 1$, that \begin{align} (xz)^{k}&=&(a+1)(ab^2+1)\\ &=&a^2b^2+a+ab^2+1\\ &\leq&b^2(a^2+a+a+1)\\ &=&b^2(a+1)^2. \end{align}

Now we note that

$$(a+1)^2=a^2+2a+1\leq a^2+2a^2+a^2=4a^2,$$

whence we see that

$$(a+1)^{2}\leq 4a^{2}\leq k^{k/2}a^{\frac{k-2}{2}}$$

since $k\geq 50$. Therefore we have by \eqref{*} that

$$(xy)^{k}\leq b^{2}(a+1)^{2}\leq b^{2}(k^{k}a^{k-2})^{1/2}\leq b^{3},$$

as desired

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