We have given the Diophantine equation $(1) \;\; p^2 + q^2 + 49r^2 = 9k^2 - 101$, where $p,q,r$ are primes and $k$ is a positive integer. If $p,q,r$ are different from 3, then $p^2 \equiv q^2 \equiv r^2 \equiv 1 \pmod{3}$, which according to equation (1) implies $3 \mid p^2 + q^2 + 49r^2$, yielding $3|101$ by equation (1). This contradiction means $p$, $q$ or $r$ equals 3. Assume $p=3$. Then by equation (1) $(2) \;\; q^2 + 49r^2 = 9k^2 - 110$, yielding $3 \mid q^2 + r^2 + 2$, which again implies $q=3$ or $r=3$. Inserting $q=3$ in equation (2) result in $(3) \;\; 9k^2 = 7(7k^2 + 17)$, yielding $7 | k$, which according to equation (3) give us $7 | 17$. Hence $q \neq 3$ by contradiction. Likewise, since $p$ and $q$ are symmetric in equation (1), $q=3$ will result in the same contradiction (i.e. $7 |17$). Consequently $r=3$, which inserted in equation (2) result in $(3k -q)(3k + q) = 19 \cdot 29$, yielding ${\textstyle (3k - q,3k + q) = (d,\frac{551}{d}), d \in \{1,19\}}$, i.e. ${\textstyle (4) \;\; 6k = \frac{551}{d} + d}$, ${\textstyle (5) \;\; 2q = \frac{551}{d} - d}$. If $d=1$, then by formula (5) we obtain $q=275$, which is not a prime. Therefore $d=19$, which according to formulas (4)-(5) yields $k=8$ and $q=5$. Conclusion:. The Diophantine equation (1) has exactly two solutions, namely $(p,q,r,k) = (3,5,3,8), (5,3,3,8)$.

parmenides51 wrote: Find all solutions of the equation $$p ^ 2 + q ^ 2 + 49r ^ 2 = 9k ^ 2-101$$with $ p$, $q$ and $r$ positive prime numbers and $k$ a positive integer. $mod3$ gives that two of $p,q,r$ is equal to $3$ First case:if $p=q=3$ then we have: $49r^2=9k^2-119$ and we close $LHS$ in consecutive square. Second case:if $p=r=3$ and we do the same.