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It is undecidable. If you could solve this, you could also solve Hilbert's 10th problem. Suppose we have an algorithm solving your problem for all $n$. Given a polynomial $p\in\mathbb[x_1,\dots,x_n]$, …

Concerning the planimetry problem: Having a suitable background, it is not hard to produce many puzzles like this. For example, here is one more pair of points on that mysterious circle: intersect a c …

The sum is zero for all $k<p^2-1$. Assume that $k$ is a multiple of $p-1$ and $k<p^2-1$. Divide all matrices into classes of the form $\{A,A+1,A+2,\dots,A+(p-1)\}$ . Summimg the $k$th powers over su …

The well-known Liouville theorem asserts that an irrational algebraic number $\alpha$ cannot have too good rational approximations, namely $|\alpha-p/q|\ge C(\alpha)/q^k$ where $k$ is the degree of $\ …

There are such functions. Moreover any diffeomorphism $f_0:\mathbb R\to\mathbb R$ can be approximated by such $f$. For the sake of simplicity I assume that $f_0'\ge 2$ everywhere. Enumerate the ratio …

The answer is infinity for $n>2$. Suppose that there is an $i$ such that $T^i(x)=0$ for all integer vectors $x$. Then the same follows for all rational vectors by homogenuity, and then for all real v …

No, being large component-wise is not enough. Consider the system $$ \begin{cases} 2x+y+z = b_1 \\ x+2y+z=b_2 \end{cases} $$ If $x,y,z\ge 0$, then obviously $b_2\le 2b_1$. So, for $b=(N,3N)$ there are …

Here is a proof that $|g(n)|\le 1$ for all but finitely many $n$. You can extract an explicit bound for $n$ from the argument and check the smaller values by hand. If $f(n)=n^{3/2}$ without the floor …