If $ p=3k$ since $ p$ is prime, $ p=3$ $ (p+q)^2>0$ then $ 27>q^5$ with $ q$ prime, this is not possible If $ p=3k+2$ Looking the equation $ mod 3$ we find no solutions if $ p=3k+1$ we find that $ q=3l$ since it is a prime number, $ q=3$ Replacing into the original equation: $ p^3-p^2-6p-252=0= (p-7)(p^2+6p+36)$ whose only natural solution is $ p=7$ So the only solution to the problem is $ (p,q)=(7,3)$

When I saw the paper of the olympiad I got the feeling I had seen a similar problem before... Unfortunately the problem is in fact not original, it was taken from the 1997 Russian Olympiad! You can verify it appears in the book Mathematical Contests 1996-97...

My solution is quite similar : If $ p = 3$, we don't have solutions, then we prove the lemma: All prime $ p \ge 5$ is of the form $ 6k + 1$ or $ 6k + 5$, $ k \in \{0,1,...,\}$. If $ p = 6k + 5$, the equation modulo $ 3$ is: $ 2 - q^5\equiv 1 + q + q^2(mod. \ 3) \implies 3|q + q^2 + q^5 - 1$, we don't have solutions, because $ q + q^2 + q^5 - 1 \not \equiv 0 (mod. \ 3)$. If $ p = 6k + 1\implies 3|2q + q^2 + q^5 \implies q\equiv0 (mod. \ 3) \implies q = 3$. Finally we have the unique solution: $ (7,3)$.

makmat wrote: My solution is quite similar : If $ p = 3$, we don't have solutions, then we prove the lemma: All prime $ p \ge 5$ is of the form $ 6k + 1$ or $ 6k + 5$, $ k \in \{0,1,...,\}$. If $ p = 6k + 5$, the equation modulo $ 3$ is: $ 2 - q^5\equiv 1 + q + q^2(mod. \ 3) \implies 3|q + q^2 + q^5 - 1$, we don't have solutions, because $ q + q^2 + q^5 - 1 \not \equiv 0 (mod. \ 3)$. If $ p = 6k + 1\implies 3|2q + q^2 + q^5 \implies q\equiv0 (mod. \ 3) \implies q = 3$. Finally we have the unique solution: $ (7,3)$. I think you haven't prove that $ p=2$ is not a solution. Eventhough, it is obvious

TFPNH wrote: makmat wrote: My solution is quite similar : If $ p = 3$, we don't have solutions, then we prove the lemma: All prime $ p \ge 5$ is of the form $ 6k + 1$ or $ 6k + 5$, $ k \in \{0,1,...,\}$. If $ p = 6k + 5$, the equation modulo $ 3$ is: $ 2 - q^5\equiv 1 + q + q^2(mod. \ 3) \implies 3|q + q^2 + q^5 - 1$, we don't have solutions, because $ q + q^2 + q^5 - 1 \not \equiv 0 (mod. \ 3)$. If $ p = 6k + 1\implies 3|2q + q^2 + q^5 \implies q\equiv0 (mod. \ 3) \implies q = 3$. Finally we have the unique solution: $ (7,3)$. I think you haven't prove that $ p = 2$ is not a solution. Eventhough, it is obvious That's correct, but of the equation we obtain that $ p>q$, so the minium value for $ p$ is $ 3$.

I know other problem that is similar: Find all primes such that $ p+q=(p-q)^3$, obviously it's easier than the other equation

this problem is definitely not new, it's rather well-known. anyway, this problem is too easy for the sixth problem i think.

I have a different solution (not as pretty as the others have shown): If $ p \equiv a \pmod q$ then $ p^3 - q^5 \equiv a^3 \pmod q$ and $ (p + q)^2 \equiv a^2 \pmod q$, therefore $ a \equiv 1 \pmod q$ since $ p > q$ hence $ a \neq 0$ and $ (a,q) = 1$. Writing $ p = kq + 1$ we have $ p^3 - q^5 \equiv 3kq + 1 \pmod {q^2}$ while $ (p + q)^2 \equiv 2(k + 1)q + 1 \pmod {q^2}$, therefore $ 3kq + 1 \equiv 2kq + 2q + 1 \pmod {q^2}$ and we get that $ k \equiv 2 \pmod q$. Now I will show that $ p < q^2$. If $ q \geq 3$ and $ p = q^2$ then $ p^3 - q^5 - (p + q)^2 = q^6 - q^5 - q^4 - 2q^3 - q^2 > q^6 - 2q^5 > q^5 > 0$. It is not hard to show that with fixed $ q$, increasing $ p$ makes the left hand side grow faster than the right hand side. After this we would have $ p < q^2$ and since $ k \equiv 2 \pmod q$ and $ p > q$ we conclude that $ p = 2q + 1$. If we let $ q \geq 5$ we will have $ p^3 - q^5 = (2q + 1)^3 - q^5 < 0$. Therefore we need only check $ q = 3 \mbox{ with } p = 2\cdot 3 + 1 = 7$ and $ q = 2$. The first leads us to a solution and then it is easy to show that $ q = 2$ yields no solution. The only solution is $ p = 7$ and $ q = 3$.

Concyclicboy wrote: Find all prime numbers $ p$ and $ q$ such that $ p^3 - q^5 = (p + q)^2$. One primm nummer greater than $ 3$, you can write how $ 3k+1$, $ 3k+2$. 1. If $ p=3k_1+1$ and $ q=3k_2+1$ then $ p^3 - q^5 \equiv 0 (mod 3)$, but $ (p+q)^2\equiv 1(mod 3)$ 2. If $ p=3k_1+2$ and $ q=3k_2+2$ It is analogous how 1. 3. If $ p=3k_1+1$ and $ q=3k_2+2$ then $ p^3 - q^5 \equiv -4, -1, -7 (mod 9)$ ; $ (p+q)^2\equiv 0(mod 9)$ 4. $ p=3k_1+2$vand $ q=3k_2+1$ then $ p^3 - q^5 \equiv 4,1,7(mod 9)$ ; $ (p+q)^2\equiv 0(mod 9)$ Now it is easy to see that $ q = 3 \Rightarrow p=7$ Fitim

Very easy problem for CentroAmerican.

For $ q\equiv_31$, by the Fermat's Little Theorem we have $ p-1\equiv_3(p+1)^2\Leftrightarrow p(p+1)\equiv_31$, so $ p\equiv_31$, but then $ p(p+1)\equiv_32$, contradiction! For $ q\equiv_3-1$, by the Fermat's Little Theorem we have $ p+1\equiv_3(p-1)^2\Leftrightarrow p^2\equiv_30$, so $ p=3$, so we have $ 3^3-q^5=(3+q)^2$, and by the Trivial Inequality, $ (3+q)^2\ge0\Leftrightarrow27\ge q^5$, so $ 27\ge q^5\ge2^5=32$, contradiction! So $ q=3$ and then we have $ p^3-3^5=(p+3)^2\Leftrightarrow p^3-p^2-6p-252=0\Leftrightarrow(p-7)(p^2+6p+36)=0$, and since by the Trivial Inequality, $ p^2+6p+36>p^2+6p+9=(p+3)^2\ge0$, we have $ p=7$, and it's easy to check $ (p,q)=(7,3)$ is really a solution.

http://www.animath.fr/old/tutorat/dossier_05063sol.pdf