Diophantine inequality with four squares and one $k$th power of primes
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Let $k$ be an integer with $k\geq 3$. Define $h(k)=\left[{\frac{k+1}{2}}\right]$, $\sigma(k)=\min\left(2^{h(k)-1},\,{\frac{1}{2}}h(k)(h(k)+1)\right)$. Suppose that $\lambda_1,\ldots,\lambda_5$ are non-zero real numbers, not all of the same sign, satisfying that $\frac{\lambda_1}{\lambda_2}$ is irrational. Then for any given real number $\eta$ and $\varepsilon>0$, the inequality \begin{align*} |\lambda_1p^2_1+\lambda_2p_2^2+\lambda_3p_3^2+\lambda_4p_4^2+\lambda_5p_5^k+\eta|<(\max_{1\leq j\leq 5} p_j)^{-\frac{3}{20\sigma(k)}+\varepsilon } \end{align*} has infinitely many solutions in prime variables $p_1,\ldots,p_5$. This gives an improvement of the recent results.
Abstract
Let $k$ be an integer with $k\geq 3$. Define $h(k)=\left[{\frac{k+1}{2}}\right]$, $\sigma(k)=\min\left(2^{h(k)-1},\,{\frac{1}{2}}h(k)(h(k)+1)\right)$. Suppose that $\lambda_1,\ldots,\lambda_5$ are non-zero real numbers, not all of the same sign, satisfying that $\frac{\lambda_1}{\lambda_2}$ is irrational. Then for any given real number $\eta$ and $\varepsilon>0$, the inequality \begin{align*} |\lambda_1p^2_1+\lambda_2p_2^2+\lambda_3p_3^2+\lambda_4p_4^2+\lambda_5p_5^k+\eta|<(\max_{1\leq j\leq 5} p_j)^{-\frac{3}{20\sigma(k)}+\varepsilon } \end{align*} has infinitely many solutions in prime variables $p_1,\ldots,p_5$. This gives an improvement of the recent results.
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