Diophantine Equations: Factorization

Diophantine equations are equations that are solved in integer numbers. We can solve some diophantine equations by factoring in one side of the equation.

Problem (42PMO, 2019)

Three prime numbers are such that their product is 103 times greater than their sum. Find all such numbers.

Solution

Answer: the only such numbers are 3, 53, and 103.

Let the numbers be $p$, $q$ and $r$. Therefore we have the following equation:

$$ pqr=103(p+q+r) $$

Notice that $103$ is prime. Since the left-hand side is divisible by 103, so should be the right-hand side and therefore one of the numbers is 103. Let this number be $p$: $p=103$. The equality now becomes

$$ qr=103+q+r $$

 

We will solve this diophantine equation by factorization:

$$ qr-q-r=103 $$

$$ qr-q-r+1=104 $$

$$ (q-1)(r-1)=104 $$

Let us consider the following cases for the factorization of $104$:

$104 = 1 \cdot 104$: therefore $(q,r)$ is $(2,105)$ or $(105, 2)$

$104 = 2 \cdot 52$: therefore $(q,r)$ is $(3,53)$ or $(53,3)$

$104 = 4 \cdot 26$: therefore $(q,r)$ is $(5,27)$ or $(27,5)$

$104 = 8 \cdot 13$: therefore $(q,r)$ is $(9,14)$ or $(14,9)$

It is not hard to see that the only pairs that represent two primes are $(3,53)$ and $(53,3)$.

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