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Let $0 < \varepsilon < 1$. A natural number $x$ is called an $\varepsilon$-square if $x = ab$, $a, b \in \mathbb{N}$ and $(1 - \varepsilon)b \le a \le b$. Denote by $f(N)$ the number of $\varepsilon …

Update. I think, now the solution to this beautiful equation is complete. The equation $$2 x^2 - y x z + 2 z^2 = y^2 - 9 y + 23$$ is equivalent to $$X^2 - f_1Y^2 = 8f_2$$ where $X = 4x - yz$, $Y = z$, …

If such $(A, B, C)$ exist, then $(\sqrt{A}, \sqrt{B}, \sqrt{C})$ are sides of a perfect cuboid (see here). Such a cuboid has not yet been found.

As you have already noticed, we may assume that $x \equiv 3 \pmod{4}$, $y \equiv 2 \pmod{4}$. Let $p \equiv 3 \pmod{4}$ be a prime divisor of $y + 1 \equiv 3 \pmod{4}$ such that $\nu_p(y+1) \equiv 1 \ …

Jacobi symbols help again. Let $x' = -x$, $z = 2z'$. In each case, we have enough information modulo 4 or 8 to calculate Jacobi symbol. case 1. $x > 0$, $z \equiv 1 \pmod{2}$ $$ x + Y^2 = (z^2 - 2x^2) …

This equation is unsolvable. Modulo 9 analysis shows that $z \ne 0 \pmod{3}$ and $z \ne \pm 1 \pmod{9}$. Let $p$ be a prime divisor of $z^2 + 3$ for which 3 is not a cubic residue. From $$z^2 + 3 = 1/ …

This is not a solution, but a promising approach to the problem. I tried to apply the method I used to solve the equation $$9x^3 + y^3 = z^2 + 3$$ The following conjecture seems to be true. Let $3z^2 …

The equation $$y^2 - x^3y + z^3 + 3 = 0$$ has no solutions. Let $t = x^3 - 2y$. Then $$x^6 - 4z^3 = t^2 + 12$$ Modulo 9 analysis shows that $t$ is not divisible by 3. Modulo 32 analysis shows that $t$ …

Let $p$ be a prime number. For every integer $m$, there are integers $u_1$, $u_2$, such that $\lvert u_1\rvert, \lvert u_2\rvert < \sqrt{p}$ and $$m \equiv u_1u_2^{-1} \pmod{p}.$$ Proof of this sta …

Let $f(x)$ be a polynomial with integer coefficients. Let $f(x) = 2^k \cdot m$ where $m$ is odd. The questions are What are the possible values of $m \mod 4$ (1, 3 or both)? I want the algorithm whic …

Let $x = \frac{a}{d}$, $y = \frac{b}{d}$, $\gcd(a, d, b) = 1$. $$b^3d = a^4 + ad^3$$ Suppose $p \mid a$ and $p \mid d$ for prime $p$. Then $b$ is not divisible by $p$. $$\nu_p(d) = \nu_p(a^4 + ad^3) …

One more way to solve the problem. Let $x = 4t + 3$. Then $$x^3 - 2 = 16t^2(4t + 9) + (108t + 25).$$ The system $$4t + 9 = a^2 \qquad 108t + 25 = b^2$$ has infinitely many solutions. It is reduced to …

Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version. The equation is unsolvable. The proof requires two theorems Theorem 1. Let $p = …