In this paper, we consider the simultaneous Pell equations $(a^2+2)x^2-y^2=2$ and $x^2-b{z}^2=1$ where $a$ is a positive integer and $b>1$ is squarefree and has at most three prime divisors. We obtain the necessary and sufficient conditions that the above simultaneous Pell equations have positive integer solutions by using only the elementary methods of factorization, congruence, the quadratic residue and fundamental properties of Lucas sequence and the associated Lucas sequence. Moreover, we prove that these simultaneous Pell equations have at most one solution in positive integers. When a solution exists, assuming the positive solutions of the Pell equation $(a^2+2)x^2-y^2=2$ are $x=x_m$ and $y=y_m$ with $m\ge 1$ odd, then the only solution of the system is given by $m=3$ or $m=5$ or $m=7$ or $m=9$.

Let $k,l,m_1,m_2$ be positive integers and let both $p$ and $q$ be odd primes such that $p^k=2^{m_1}-a^{m_2}$ and $q^l=2^{m_1}+a^{m_2}$ where $a$ is odd prime with $a\equiv 5(\mathrm{mod}8)$ and $a\not\equiv 1(\mathrm{mod}5)$. In this paper, using only the elementary methods of factorization, congruence methods and the quadratic reciprocity law, we show that the exponential Diophantine equation ${\left(\frac{q^{2l}-p^{2k}}{2}n\right)}^x+(p^k{q}^{l}n{)}^y={\left(\frac{q^{2l}+p^{2k}}{2}n\right)}^z$ has only the positive integer solution $(x,y,z)=(2,2,2)$.

Let $k,l,m_1$ and $m_2$ be positive integers and let both $p$ and $q$ be odd primes such that $p^k=2^{m_1}-a^{m_2}$ and $q^l=2^{m_1}+a^{m_2}$ where $a$ is a positive integer with $a\equiv \pm 3(\mathrm{mod}8)$. In this paper, using only the elementary methods of factorization, congruence methods and the quadratic reciprocity law, we show that Je $\acute{s}$manowicz' a conjecture holds for the following set of primitive Pythagorean numbers:

$$\frac{q^{2l}-p^{2k}}{2},p^k{q}^l,\frac{q^{2l}+p^{2k}}{2}.$$

We also prove that Je $\acute{s}$manowicz' conjecture holds for non-primitive Pythagorean numbers:

$$n\frac{q^{2l}-p^{2k}}{2},n{p}^k{q}^l,n\frac{q^{2l}+p^{2k}}{2},$$

for any positive integer $n$ if for $a=a_1{a}_2$ with $a_1\equiv 1(\mathrm{mod}8)$ not a square and $gcd(a_1,a_2)=1$, then there exists a prime divisor $P$ of $a_2$ such that $\left(\frac{a_1}{P}\right)=-1$ and $2|m_1,a\equiv 5(\mathrm{mod}8)$ or $2\text{⧸}|m_2,a\equiv 3(\mathrm{mod}8)$.