Find p, q, r different primes Can you find p, q, r, three different primes, so that: $$p^2+10\:\vdots\:qr$$
$$q^2+10\:\vdots\:pr$$
$$r^2+10\:\vdots\:pq$$
What if you change 10 with 11? Here $\vdots $ means divisible by. I reckon $a\:\vdots\:b$ is the same as writing $b|a$
The book solution that I am given is very ambiguous and I don't understand it. It starts from the hypothesis that $p<q<r$ and then says that this will result in $r|p^2+10$ and $r|q^2+10$ (which I don't understand, the second part at least). Then it says that $r|\left(q-p\right)\left(q+p\right)$ and because of the fact that $r>q-p>0$ we have that $r|p+q$ (drawing a blank here as well, I dont get this). It then says that because $p+q\:<\:2r$ (why!?) we have $p+q\:=\:r$ so then $p=2$ which will lead in a contradiction in $qr\:|\:p^2+10$.
I thought it was a typo at first and they wanted to say that we have $p\:|\:r^2+10$ and $p\:|\:q^2+10$ which I would understand, but even so, the train of thought they follow still leaves me with questions.
 A: $p < r, q < r \Rightarrow p+q < 2r$.The part you drew blank is : $r$ is a prime and $r \mid ab$ then $r \mid a$ or $r \mid b$. Since $a = q-p < q < r, r \nmid a \Rightarrow r \mid b = p+q$.Also since $r \mid p+q \Rightarrow r \leq p+q < 2r$. If $p+q = r + k, k < r$ then $r \mid r+k \Rightarrow r \mid k$, contradiction since $k< r$.Thus $k=0$, and $p+q =r$.
A: My answer is different but I still suppose it proves the required result.
I'll be using the $|$ symbol which stands for 'divides'.
See that $$pq|q^2 + 10 \implies q |q^2 + 10 \implies q|10$$
We also know that $q$ is a prime. Hence, $q$ is equal to $2$ or $5$.
Now we will use the first statement.
Assume that $q=2$.
Then, $2r|p^2+10 \implies 2|p^2$
Now, if $2|p^2$ then $2|p$. The only prime divisible by $2$ is 2 itself and hence $p=2$. Contradiction. If you look at the problem closely, it states that $p$, $q$ and $r$ are distinct.
Assume that $q=5$.
Then, $5r|p^2+10 \implies 5|p^2$
Now, if $5|p^2$ then $5|p$. The only prime divisible by $5$ is 5 itself and hence $p=5$. Contradiction. If you look at the problem closely, it states that $p$, $q$ and $r$ are distinct.
So, $q \neq 2 \land q\neq 5$. Contradiction. So no such $q$ exists.
A: The hypothesis that the numbers be prime is unnecessary. It suffices to assume they are distinct positive integers. 
Without loss of generality, $p<q<r$. Then $q\ge p+1$ and $r\ge p+2$, so $qr\ge(p+1)(p+2)=p^2+3p+2$. But $qr\mid p^2+10$, so $qr\le p^2+10$. Hence, $$p^2+3p+2\le p^2+10$$ which is $3p\le8$, which implies $p=1$ or $p=2$. 
If $p=1$, then $qr\mid11$, which is incompatible with $q\ge p+1=2$ and $r\ge p+2=3$. 
If $p=2$, then $qr\mid 14$, which is incompatible with $q\ge p+1=3$ and $r\ge p+2=4$. 
Thus there are no solutions, prime or otherwise. 
