The following proposition is well known (Dirichlet's theorem): Proposition 1: Let $\alpha$ is irrational number. For every positive integer $N$ there exist $p,q \in \mathbb{N},\, q \leq N$ such that: $|\alpha q - p| < \frac{1}{N}$. If one knows the proof of the above assertion it is not difficult to construct a proof for the following statement. Proposition 2: Let $\alpha, \, \beta$ are irrational numbers. For every possitive integer $N$, there exist $p_1,p_2,q \in \mathbb{N}, \, q \leq N $ such that: (1) $|\alpha q - p_1| < \frac{1}{\lfloor \sqrt{N} \rfloor}$. (2) $|\beta q - p_2| < \frac{1}{\lfloor \sqrt{N} \rfloor }$. Proof: Let $x_n = \{\alpha n\},\, y_n = \{\beta n\},\, n= 0,1,\ldots,N$. Then $(x_n,y_n) \in (0,1)\times (0,1)$. Divide the square $(0,1)\times (0,1)$ into at most $N$ squares with dimension $\frac{1}{\lfloor \sqrt{N} \rfloor }\times \frac{1}{\lfloor \sqrt{N} \rfloor } $ By PHP it follows that there exist $ (x_k,y_k)$ and $(x_\ell,y_\ell)$ placed in the same square. This means: $|x_k-x_\ell| < \frac{1}{\lfloor \sqrt{N} \rfloor }$ and $|y_k-y_\ell| < \frac{1}{\lfloor \sqrt{N}\rfloor }$ which implies (1) and (2) for $q=|k-l|$. $\square$ Now back to the problem's statement. Fix $N \in \mathbb{N}$ to be a perfect square. According to Proposition 2 there exist $p_1,p_2,q \in \mathbb{N}$ depending on $N$, $q\leq N$ such that: $|\sqrt{3}-\frac{p_1}{q}| < \frac{1}{q \sqrt{N}} \leq q^{-\frac{3}{2}} $, $|\sqrt{2}-\frac{p_1}{q}| < \frac{1}{q \sqrt{N}} \leq q^{-\frac{3}{2}} $, Obviously when $N$ varies, $(\frac{p_1}{q}, \frac{p_2}{q}) $ form infinitely many different pairs.