No Arabic abstract
In 2012, T. Miyazaki and A. Togb{e} gave all of the solutions of the Diophantine equations $(2am-1)^x+(2m)^y=(2am+1)^z$ and $b^x+2^y=(b+2)^z$ in positive integers $x,y,z,$ $a>1$ and $bge 5$ odd. In this paper, we propose a similar problem (which we call the shuffle variant of a Diophantine equation of Miyazaki and Togb{e}). Here we first prove that the Diophantine equation $(2am+1)^x+(2m)^y=(2am-1)^z$ has only the solutions $(a, m, x, y, z)=(2, 1, 2, 1, 3)$ and $(2,1,1,2,2)$ in positive integers $a>1,m,x,y,z$. Then using this result, we show that the Diophantine equation $b^x+2^y=(b-2)^z$ has only the solutions $(b,x, y, z)=(5, 2, 1, 3)$ and $(5,1,2,2)$ in positive integers $x,y,z$ and $b$ odd.
We show that the Diophantine equation given by X^3+ XYZ = Y^2+Z^2+5 has no integral solution. As a consequence, we show that the family of elliptic curve given by the Weierstrass equations Y^2-kXY = X^3 - (k^2+5) has no integral point.
We show that the diophantine equation $n^ell+(n+1)^ell + ...+ (n+k)^ell=(n+k+1)^ell+ ...+ (n+2k)^ell$ has no solutions in positive integers $k,n ge 1$ for all $ell ge 3$.
Let $f(x)=x^{2}(x^{2}-1)(x^{2}-2)(x^{2}-3).$ We prove that the Diophantine equation $ f(x)=2f(y)$ has no solutions in positive integers $x$ and $y$, except $(x, y)=(1, 1)$.
We study a variant of a problem considered by Dinaburg and Sinai on the statistics of the minimal solution to a linear Diophantine equation. We show that the signed ratio between the Euclidean norms of the minimal solution and the coefficient vector is uniformly distributed modulo one. We reduce the problem to an equidistribution theorem of Anton Good concerning the orbits of a point in the upper half-plane under the action of a Fuchsian group.
Let $Lambda(n)$ be the von Mangoldt function, and let $[t]$ be the integral part of real number $t$. In this note, we prove that for any $varepsilon>0$ the asymptotic formula $$ sum_{nle x} LambdaBig(Big[frac{x}{n}Big]Big) = xsum_{dge 1} frac{Lambda(d)}{d(d+1)} + O_{varepsilon}big(x^{9/19+varepsilon}big) qquad (xtoinfty)$$ holds. This improves a recent result of Bordell`es, which requires $frac{97}{203}$ in place of $frac{9}{19}$.