Find all triples $(p, q, r)$ of prime numbers such that all of the following numbers are integers $\frac{p^2 + 2q}{q+r}, \frac{q^2+9r}{r+p}, \frac{r^2+3p}{p+q}$ Proposed by Tajikistan
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Tags: number theory
13.09.2020 10:33
The only triplets of primes $(p,q,r)$ satisfying the conditions $(1) \;\; \frac{p^2 + 2q}{q + r} = F(p,q,r)\in \mathbb{N}$, $(2) \;\; \frac{q^2 + 9r}{r + p} = G(p,q,r) \in \mathbb{N}$, $(3) \;\; \frac{r^2 + 3p}{p + q} = H(p,q,r) \in \mathbb{N}$, is $(p,q,r) = (2,3,7)$. Proof: Let consider the following two cases: Case 1: $q + r$ is even. Then $2 \mid p^2$ by condition (1), i.e. $p=2$, which according to condition (1) give us $\frac{2q + 4}{q + r} = F(2,q,r) \leq 2$, yielding $q = r - 4$ or $r=2$. If $r=2$, then $q=2$ (since $q+r$ is even), which implies $H(p,q,r) = H(2,2,2) = 2,5$. This contradiction means $r \neq 2$, Hence $q = r - 4$ (implying $r \geq 7$) , which according to condition (3) give us $G(p,q,r) = G(2,r-4,r) = \frac{r^2 + 3 \cdot 2}{2 + (r - 4)} = \frac{r^2 + 6}{r - 2} = r + 2 + \frac{10}{r - 2}$, yielding $r - 2 \mid 5$ (since $7 \leq r$ is an odd prime). Hence $r=7$ is the only possibility, which give us $q = r - 4 = 7 - 4 = 3$. Thus $(p,q,r) = (2,3,7)$, yielding $F(2,3,7) = 1, G(2,3,7)=8, H(2,3,7)=11$. In other words, the only triplets $(p,q,r)$ of primes satisfying conditions (1)-(3) is $( p,q,r) = (2,3,7)$. Case 2: $q + r$ is odd. Then $q=2$ or $r=2$. Assume $r=2$. Then $q$ is odd (since $q+r$ is odd). Moreover by condition (3) $\frac{3p + 4}{p + q} = H(p,q,2) < 3$. yielding $q = 2(p + 2)$ (which is impossible since $q$ is odd) or $p = 2(q - 2)$, yielding $p=2$ and $q=3$, i.e. $( p,q,r) = (2,3,2)$. But $G(2,3,2) = 6,75$, which means $r \neq 2$. Therefore $q=2$ and $r$ is odd. Hence by condition (2) $(4) \;\; p + r \mid 9r + 4$. Now $9r + 4$ is odd since $r$ is odd. Consequently $p + r$ is odd by condition (4), implying $p$ is even (since $r$ is odd). Hence $p=2$, which according to condition (3) implies $4 \mid r^2 + 6$. This contradiction completes the proof.