The equation $175a + 11ab + bc = abc$

Question

Consider all the triples $(a, b, c)$ of prime numbers that satisfy the equation $$175a + 11ab + bc = abc\ .$$ Compute the sum of all possible values of $c$ in such triples.

I could only get to the point where $a$ is equal to $b$ or $c$; $b$ is equal to $5$ or $7$ or $a$; $c > 11$. Сan you please help me with further solution?

Answer 1

The only solutions are $(47,5,47),(37,7,37)$ and $(2,2,197)$ hence the required sum is $47+37+197 = 281$.

Given $175a + 11ab + bc = abc$, LHS is divisible by $a$ iff $a|bc$ and since $a,b,c$ are primes, we must have $a = b$ or $a = c$.

Case 1: If $a = b$ the $$ 175a + 11a^2 + ac = a^2 c $$

$$ c(a-1) = 175 + 11a = 186 + 11(a-1) $$ Hence $(a-1)|186$ or $a-1$ is one of the divisors of $186$ or $a-1 \in (1,2,3,6,31,62,93,186)$ or $a \in (2,3,4,7,32,63,94,187)$. Out of these, the only possible prime value of $a$ are $(2,3,7)$.

If $a = 2$ then $c = 197$ which is a prime. If $a = 3$ then $c = 104$ which is composite. If $a = 7$ then $c = 42$ which is composite. Hence the only solution in this case is $(a,b,c) = (2,2,197)$

Case 2: if $a=c$ then $$175a + 11ab + ab = a^2 b$$

$$ b(a - 12) = 175 = 5^2 7 \tag 1 $$

Since $b$ is prime, the possible values of $b$ are $5$ or $7$. If $b = 5$ then $a = 47$ which is prime and if $b = 7$, then $a = 37$ which is also prime. Hence the solutions in this case are $(47,5,47)$ and $(37,7,37)$

Answer 2

You already have $a=b$ or $a=c$ and there is the justification for it already above. Let's make two exclusive classes with $a=b$ (and may or may not equal $c$ as well) or $a=c \neq b$.

$$ 175a+11ab+bc=abc\\ 175a=(ac-c-11a)b $$

This is the $b = 5,7,a$. That is the second condition you had established that you knew in the question, but I'm including it for completeness.

So now we have 3 mutually exclusive cases

• $a=b$
• $a=c \neq b=5$
• $a=c \neq b=7$

Third case:

$$ 175a+77a+7a=7a^2\\ (175+77+7)a=7a^2\\ 259=7a\\ a=37\\ (a,b,c) = (37,7,37)\\ $$

Second case:

$$ 175a+55a+5a=5a^2\\ (175+55+5)a=5a^2\\ 235=5a\\ a=47\\ (a,b,c) = (47,5,47)\\ $$

First case:

$$ 175a+11a^2+ac=a^2c\\ 175+11a+c=ac\\ 175=ac-11a-c\\ 175=(a-1)(c-11)-11\\ 186=(a-1)(c-11)\\ 2*3*31=(a-1)(c-11)\\ $$

$a$ must be such that $a-1$ is a factor of $186$ so for choosing that factor to be $1,2,3,6,62,93,186$ respectively we get $a=2,3,4,7,63,94,187$ of which $2,3,7$ are kept because $a$ must be prime. In those cases $c=197,104,42$. The demand on $c$ being prime too means only $(2,2,197)$ remains from this case.

$$ c_{sum} = 37+47+197 $$