It is easy to see that (r+1)/(r+5)=q/p. So r+1=q and r+5 = p. So r=2 , q=3 and p=7.

Actually $ r + 1 = q$ and $ r + 5 = p$ isn't necessary. Also, $ p = 3,q = 2,r = 7$ is another solution which doesn't satisfy your condition.

Solution below

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As suggested by Taidoda-X, note that $ \dfrac{p}{q}=\dfrac{r+5}{r+1}$. Euclid's algorithm immediately yields that $ \lambda(r)=(r+5,r+1)=(4,r+1)$. Hence, $ \lambda(r)\in\{1;2;4\}$. Note also that $ p>q$. The rest is little more than casework: $ \lambda(r)=1$: This readily implies $ r+1=q$, whence the solution $ (p,q,r)=(7,3,2)$ $ \lambda(r)=2$: We get $ 2p=r+5$ and $ 2q=r+1$. Thus, additionning and subtracting the two equalities, $ p+q=r+3$ and $ p-q=2$. Suppose $ p,q,r\geq5$. Hence, $ p+q\equiv 0,2\mbox{ mod }6$ and $ r+3\equiv 2,4\mbox{ mod }6$, implying that $ p\equiv q\equiv 1\mbox{ mod }6$, contradicting the fact that $ p$ and $ q$ are twin primes. Checking the remaining cases, we find the solution $ (p,q,r)=(5,3,5)$. $ \lambda(r)=4$: Subtracting the equalities $ 4p=r+5$ and $ 4q=r+1$, we find that $ p-q=1$, which readility implies the solution $ (p,q,r)=(3,2,7)$.

p(r+1)=q(r+5) Now, take cases for r=2,3,6k-1,6k+1 Case(1): r=2 p=7,q=3 (7,3,2) Case(2): r=3 p/q=9/4. So, no solution Case(3): r=6k-1 p*3k = q*(3k+2) 3k+2 is not divisible by 3 => 3 is a divisor of q => q=3 => kp=3k+2 => k(p-3)=2 (i) k=2 => p=4 contradiction (ii)k=1 => p=5 => r=5 (5,3,5) Case(4): r=6k+1 p*(3k+1)=3q*(k+1) 3k+1 is not divisible by 3 => 3 is a divisor of p => p=3 => 3k+1=kq+q => k(3-q)=(q-1). k>0 => q = 2 => k=1 => r =7 (3,2,7) Solutions: (7,3,2),(5,3,5) and (3,2,7)

I'am not sure is this correct.... We easily see that r=(5q-p)/(p-q)=-5+(4p)/(p-q) As r is integer, we see that 4p=k(p-q), k is positive integer. 4p=2*2*p p-q<p We see that p-q=1 or p-q=2 or p-q=4 If p-q=1, then p=3, q=2, r=7 If p-q=2, then p=5,q=3, r=5 or p=6k+1, q=6k-1. It follows that r=12k-3. Then r is not prime, contradiction. If p-q=4, then p=7,q=3,r=2 or p=6k+5, q=6k+1. Then r=6k, r is not prime, again contradiction. Solutions: (3,2,7), (5,3,5), (7,3,2).

My solution..... p/q-4/(r+1)=1 [p(r+1) - 4q] / q(r+1) = 1 p(r+1) - 4q = q(r+1) pr+p-4q = qr+q pr - qr = 4q + q - p r(p-q) = 4q + q - p r = [4q + (q - p)] / [p-q] r = [4q - (p - q)] / p - q r = 4q / (p-q) - 1 p-q|4q p-q = 1 ; p-q = 2 ; p-q = 4 For the first and third case the solution [7,3,2] ; but for second case p,q,r isn't simple number.....

naye,will you post your full solution?I am curious.

An easier solution: $r+1=\frac {4q} {p-q},gcd(q,p-q)=1\implies p-q=1,2,4$ #$1.p-q=1,p=3,q=2,r=7$ #$2.p-q=2,q=6n-1,$if $p>3,$ and then $r=3(4n-1),$so $p=5,q=3,r=5$ #$p-q=4,r+1=q\implies q=3,r=2,p=7$ That's all we need.

Can you explain your solution more clearly?

$\# .$If $p-q=,$it has the only solution $p=3,q=2$ because then they are of different pairity,so the smaller one is even i.e. $q=2$.Then $r=7$ $\#2.$If $p-q=2$ then it is well known that every prime greater than $3$ is of the form $6n\pm 1$ Using this,$p\equiv 1\mod 6,q\equiv -1\mod 6$ but then $r+1=2q\implies r=2(6q-1)-1=3(2q-1)$ which is not a prime.So then the only possible value is $p=5,q=3$ $\#3.p-q=4,r+1=q$ so then $r=2,q=3$ is the only solution

p,q,r-prime pr+p-4=qr+q =>pr+p-qr-q=4 =>p(r+1)-q(r+1)=4; (r+1)(p-q)=4 System 1)r+1=1 p-q=4 =>r=0 it’s not true 2)r+1=4 p-q=1 => p-q=1 it’s not true 3)r+1=2 p-q=2=>r=1;p=2+q

Above is not true because $(r+1)(p-q)=4q$ Solution:Let $r\equiv{1}(mod3)$ $\implies$ $q=3$$\implies$ $12=(p-q)(r+1)$ , done ! Let $r\equiv{2}(mod3)$ $\implies$ $p=3$ $\implies$ $3(r+1)=q(r+5)$ $\implies$ $q<3$ , done! Let $r=3$ $\implies$ $p=2q$ , absurd.

Offer to find all the solutions of this equation, and not just simple numbers. Need to write a formula describing the solutions of this equation.

individ wrote: Offer to find all the solutions of this equation, and not just simple numbers. Need to write a formula describing the solutions of this equation. In each case, 7 years old child can find solutions.

Ahiles wrote: Find all prime numbers $ p,q,r$, such that $ \frac{p}{q}-\frac{4}{r+1}=1$

Ahiles wrote: Find all prime numbers $ p,q,r$, such that $ \frac{p}{q}-\frac{4}{r+1}=1$ Another way... Obvious $p=q$ no solutions so...$(p-q,q)=1$ $\dfrac{p-q}{q}=\dfrac{4}{r+1}$ (1) So $p-q=1$ or $p-q=2$ or $p-q=4$ If $p-q=1$ then obvious $(p,q,r)=(3,2,7)$ If $p-q=4$ from (1) $q=r+1$ so obvious $(p,q,r)=(7,3,2)$ If $p-q=2$ from $(1)$ $r+1=2q$ Because $p-q=2$ if $q=3$ we have the solution $(p,q,r)=(5,3,5)$ but if $q>3$ then p>3$ and because $p-q=2$ must $p=1mod6$ , q=5mod6$ so from $r+1=2q$ we have $r=9mod6$ so only if $r=3$ then $q=2$ and $p=4$ no solution... $(p,q,r)=(3,2,7),(7,3,2),(5,3,5)$

My solution: Rearrange the condition as $\frac{p}{q}=\frac{r+5}{r+1}$. Observe that $\gcd{(r+5, r+1)}=\gcd{(4, r+1)}=x$. Case 1: $x=1$. Then we must have $r$ even, so $r=2$. Then $p=7, q=3$ yielding the solution $(7, 3, 2)$. Case 2: $x=2$. This is only true if $r=4a+1$ for some integer $a \ge 0$. Then we have $p=2a+3, q=2a+1$ because $\gcd{(2a+3, 2a+1)}=1$. Suppose we have a solution in the form $(2a+3, 2a+1, 4a+1)$. Then either $a \equiv 2 \pmod{3}$ or $a=1$. If $a \equiv 2 \pmod{3}$ then $4a+1$ is not prime, so $a=1$. This gives us the solution $(5, 3, 5)$. Case 3: $x=4$. This is only true if $r=4a+3$ for some integer $a \ge 0$. Hence, $p=2a+2, q=2a+1$. The only primes that differ by $1$ are $2$ and $3$, so $a=1$ and we have the solution $(3, 2, 7)$. So the solutions are $(p, q, r)=(3, 2, 7), (5, 3, 5), (7, 3, 2)$. We are done.

The given equation can be rearranged into the below form: $4q = (p-q)(r+1)$ $Case 1: 4|(p-q)$ then we have $q = ((p-q)/4)(r+1)$ $=> (p-q)/4 = 1$ and $q = r + 1$ $=> r = 2, q = 3$ and $p = 7$ $Case 2: 4|(r+1)$ then we have $q = (p-q)((r+1)/4)$ $=> (p-q) = 1$ and $q = (r + 1)/4$ $=> p = q + 1 => q = 2, p = 3$ and $r = 7$ note that if $(r+1)/4 = 1$, then $q = (p-q) => p = 2q$ which is a contradiction. $Case 3: 2|(p-q)$ and $2|(r+1)$ then we have $q = ((p-q)/2)((r+1)/2)$ $=> (p-q)/2 = 1$ and $q = (r+1)/2$ $=> p = q + 2$ and $r = 2q - 1$ We have that exactly one of $q, q + 1, q + 2$ is a multiple of $3$. $q + 1$ cannot be a multiple of $3$ since $q + 1 = 3k => q = 3k - 1$. Since $r = 2q - 1$ is prime, then we have $2(3k - 1) - 1 = 3(2k-1)$ is a prime. $=> k = 1 => q = 2 => p = 4$ contradiction. Also, $q + 2$ cannot be a multiple of $3$ since, $q + 2 = 3k => p = 3k => k = 1 => q = 1$ contradiction. So, $q = 3k$ $=> k = 1 => q = 3, p = 5$ and $r = 5$ Thus we have the following solutions: $(7, 3, 2), (3, 2, 7), (5, 3, 5)$

Note that \[\frac{p}{q} - \frac{4}{r+1} = 1 \iff (p-q)(r+1) =4q.\]We must have $p>q$ and $p-q\mid 4q$. Because $\gcd (p-q,q)=1$, so we have $p-q\mid 4$. If $p-q=1$, we have $(p,q,r) = (3,2,7)$. If $p-q=2\iff p=q+2$, we also have $r=2q-1$. Note that $(p,q,r) = (5,3,5)$ is solution. Consider when $q>3$. Let $q= 6k+1\implies p= 6k+3$ or $q=6k+3\implies r=12k + 9$ for some positive integer $k$, but this contradiction $p,r$ is prime numbers. If $p-q=4\iff p =q+4$, we alsp have $r = q-1$. Then, $(p,q,r) = (7,3,2)$. Therefore, the solution : $(p,q,r) = (3,2,7),(5,3,5),(7,3,2)$.

Multiplying by $q(r+1)$ gives $p(r+1)-4q=q(r+1)$, or $pr-4q+p=qr+q$, equivalent to $pr+p=qr+5q$ after simplification. This means $p(r+1)=q(r+5)$, or $\frac{p}{q}=\frac{r+5}{r+1}$. Now we take cases. From Euclidean Algorithm, the maximum value of $\gcd(r+5,r+1)$ is $4$, attained with $r=4k+3$, giving $\frac{p}{q}=\frac{4k+8}{4k+4}=\frac{k+2}{k+1}$. This would mean $p=k+2, q=k+1$, only possible with $k=1 \implies (3,2,7)$. We cannot have $\gcd(r+5, r+1) = 3$, as $r+5 \equiv 0 \pmod 3$ means $r \equiv 1 \pmod 3$ and $r+1 \equiv 2 \pmod 3$. If $\gcd(r+5,r+1)=2$, then $r=4k+1$, so $\frac{p}{q}=\frac{2k+3}{2k+1}$. These two have GCD $1$, so $p=2k+3, q=2k+1, r=4k+1$. Assume $p, q, r \neq 3$. Then the second equation gives $k \equiv 2 \pmod 3$, but this is not possible due to the third equation. If $p=3$, then $k=0$, which does not work. If $q=3$, then $k=1$, which gives $(5,3,5)$. If $r=3$, then $k=\frac{1}{2}$, which does not work. Lastly, we could have $\gcd(r+5, r+1)=1$, which means $p=r+5, q=r+1$. Clearly $r$ then must be even, so $r=2$, which gives $(7,3,2)$. Our solutions are $\boxed{(3,2,7),(5,3,5),(7,3,2)}$.

daniel73 wrote: Solution below

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Thanks! That second hint really helped me, then it was just casework. My sols were $(p, q, r) = (3, 5, 5), (3, 2, 7), and (2, 3, 7).$