Lemma $ a$, $ b$ and $ k$ is positive integer and $ p$ is a prime number satisfying $ p^{k}>b$. Then $ v_{p}(ap^{k}+b)=v_{p}(ap^{k})+v_{p}(b)$. Proof $ v_{p}(ap^{k}+b)-v_{p}(ap^{k})$ is equal to $ p$-exponent of $ (ap^{k}+1)(\cdots)(ap^{k}+b)$. Because $ p$-exponent of $ b$ is less than $ k$, $ ap^{k}+c$ and $ c(\le b)$ have same $ p$-exponent. Therefore $ v_{p}(ap^{k}+b)-v_{p}(ap^{k})=v_{p}(b)$. Let's make a infinite sequence $ \{x_{i}=p_{1}^{e_{i,1}}\cdots p_{k}^{e_{i,k}}\}$ such that $ x_{i}< p_{j}^{e_{i+1,j}}$ for all $ 1\le j\le k$. and $ x_{1}+\cdots+x_{m}=y_{m}$ There are only $ d^{k}$ possible case of $ (v_{p_{1}}(y_{m}),\cdots,v_{p_{k}}(y_{m}))\pmod d$ So there is a infinite subsequence of $ y_{m}$s $ \{y_{i_{m}}\}_{m}$ such that above vector is same. Then by Lemma, $ v_{p_{j}}(y_{i_{a}}-y_{i_{b}}) \equiv 0 \pmod d$ $ (1\le j\le k, a>b)$ These $ y_{i_{a}}-y_{i_{b}}$ are all can be a solution.