Estonian Math Competitions

Subtracting the second equation from the first one gives $3p+2q-3s=0$ which implies $2q=3(s-p)$. Thus $2q$ is divisible by $3$. As both $2$ and $q$ are primes, this implies $q=3$. Substituting $q=3$ into the initial system of equations and simplifying gives $$\{\begin{array}{l}6p+5r+3s=115,\\ 3p+5r+6s=121.\end{array}(1)$$ If $r$ and $s$ were both odd then also $5r$ and $3s$ would be odd, whence $5r+3s$ would be even. As $6p$ is even, too, the left-hand side of the first equation of system (1) would be even and could not equal the right-hand side $115$. Thus one of $r$ and $s$ is even, i.e., $r=2$ or $s=2$. Analogously if $p$ and $r$ were both odd then the second equation of system (1) would give a contradiction; hence $p=2$ or $r=2$. Consequently, if $r\ne 2$ then $p=s=2$, but substituting $p=s$ into system (1) and subtracting the second equation from the first one gives $0=-6$. The contradiction shows that $r=2$. Substituting $r=2$ into system (1) and simplifying gives $$\{\begin{array}{l}6p+3s=105,\\ 3p+6s=111.\end{array}$$ Solving this equation gives $p=11$ and $s=13$.

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Alternative solution.

As in Solution 1 we find $q=3$ and obtain system (1). Subtracting the first equation from the second one in system (1) gives $3s-3p=6$, implying $s=p+2$. The least pairs of prime numbers with difference $2$ are $(3,5)$, $(5,7)$ and $(11,13)$. The next such pair $(17,19)$ gives $6p+3s>115$, implying that no more pairs are suitable. Substituting the three pairs one by one into system (1) and simplifying each time gives $5r=82$, $5r=64$ and $5r=10$, respectively. Only the last alternative leads to an integral solution $r=2$, which is a prime, too. Consequently, the only possibility is $(p,q,r,s)=(11,3,2,13)$.