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My head is turning into a uniform gel of random thoughts! I cannot see a proof or find a counterexample to the following:

Conjecture: Let integer $x$ be expressed as $a_3 \, b^3 + a_2 \, b^2 + a_1 \, b + a_0$ in base $b$. If $a_3 \ge b/4$ then $\lfloor \sqrt{x} \rfloor$ is independent of $a_0$.

Any help is appreciated.

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I am afraid it does depend!

Take $b=2$ and $x=8$, in which case $x=1\cdot 2^3+0\cdot 2^2+0\cdot 2+0\cdot 1$, i.e., $a_0=a_1=a_2=0$ and $a_3=1$. Clearly the condition $a_3\ge b/4$ is satisfied since $a_3=1> b/4=1/2$ and $\lfloor\sqrt{x}\rfloor=2$. On the other hand, if $a_0$ becomes equal to $1$, then $x$ becomes $9$ and $\lfloor\sqrt{9}\rfloor=3$.

Also try $x=3024$, with $b=10$, where $a_3=3>10/4$, $\lfloor\sqrt{x}\rfloor=54$ and $\lfloor\sqrt{x+1}\rfloor=55$.

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  • $\begingroup$ Of course. Thank you. $\endgroup$ – user114628 Dec 15 '13 at 0:32
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Take $b=10$ and e.g. $x=3135 \ (= 56^2-1)$. Then $\lfloor\sqrt x\rfloor=55$ but $\lfloor\sqrt{3136}\rfloor=56$.

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