8 problems

Non-polynomiality conjecture. No q-deformed cubic irrational can satisfy a cubic equation of order $3$ with polynomial coefficients.

Let $p(x)=x^3+2x^2+10x-20$, let $x_5=1+22/60+7/60^2+42/60^3+33/60^4+4/60^5$, and let $k$ be an integer. Fibonacci's selected-checks conjecture. Fibonacci may have calculated only ……

Let $p(x)=x^3+2x^2+10x-20$, let $x_5=1+22/60+7/60^2+42/60^3+33/60^4+4/60^5$, and let $k$ be an integer. Fibonacci's computational-effort conjecture. Fibonacci may have avoided calc…

For a positive integer $n$, let $r_k(n)$ denote the number of representations of $n$ as a sum of $k$ positive integral $k$-th powers, and let $\mathscr{C}$ be the set of integers r…

For integers $t\geq 3$, define … The numerical evidence suggests that $f_3=0$ and $f_t>0$ for $t\geq 4$. Positivity conjecture. For integers $t\geq 4$, $f_t>0$. On the other hand,…

Hooley's conjecture. A similar phenomenon should hold in a circle method approach to counting integral solutions of

Let $r=2d$ with $1\leq d\leq 50$, let $p\geq 5$ be prime, and consider the equation … Here $d,x,z$ are integers. Alternating-cube perfect-power conjecture. The possible integer sol…

An even perfect number is a positive integer equal to the sum of its proper positive divisors. The claim concerns such numbers other than $6$. Three-cube conjecture. Any even perfe…