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Find all pairs of prime numbers $p, q$ such that $p+q = 18(p−q)$.

It is clear that $p-q$ must be an even number since if we consider $q$ as $2$, we won't get any solution. So any pair of odd prime does the work. I got $p=19$ and $q=17$ as one pair, considering the fact that $p+q$ must be a multiple of $18$. So considering numbers $18\cdot 2$, $18\cdot4$ , $18\cdot6$ ... as $p+q$ and then testing whether such pair of odd prime exist or not.

Is this the right approach since I am stuck up?

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$p+q = 18p - 18q$

$17p = 19q$

Therefore $p = 19n$ and $q=17n$.

Then $p$ and $q$ are prime only when $n=1$.

Hence the only solution in primes is $(p,q) = (19,17)$

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  • $\begingroup$ Thanks for the solution. Logic is right. $\endgroup$ – Subhra Mazumdar Apr 13 '16 at 2:08
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You can write this as $$17p=19q$$ $$\implies \frac{p}{q}=\frac{19}{17}$$ But since $p,q$ are prime, the only possible solution is $p=19,q=17 $.

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I don't think yours is the best approach.

Observe that $$ p + q = 18(p-q) \iff 17p = 19q $$

So $ p = 19, q = 17 $ satisfy this, but if $ p \neq 19 $, then $17p $ must be divisible by $ 19 $ and since $ \gcd(17,19) = 1$, so must be $ p $. This proves that $(17,19)$ is the only such pair

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$\textbf{Hint:}$ Rearranging leads to $17p=19q$. This should suggest more to you. Note that $17$ and $19$ are prime.

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