old http://www.artofproblemsolving.com/community/q1_%22p%5E3%20-%20q%5E5%20%3D%20(p%20%2B%20q)%5E2%22

$(p+q)^2 = p^3 - q^5 $ $q \ge p \implies 0 > p^3-q^5 = (p+q)^2 $ So $p>q$ $(p+q)^2\equiv p^2\pmod q$ and $p^3-q^5\equiv p^3\pmod q \implies p^3\equiv p^2\pmod q \implies q|p^3 - p^2 = p^2(p-1) \implies q|p-1$ $(p+q)^2 \equiv q^2 \pmod p$ and $p^3-q^5 \equiv -q^5 \pmod p \implies q^2 \equiv -q^5 \pmod p \implies p|q^2+q^5=q^2(q^3+1) \implies p|q^3+1$ $ p|(q+1)(q^2-q+1) \implies p|q+1$ or $p|q^2-q+1$ $i-) p|q+1 \implies q+1 \ge p$ and $q|p-1 \implies p-1 \ge q \implies p-1 \ge q \ge p-1 \implies p-1=q \implies q = 2 $ and $ p = 3$ but $3^3-2^5 < 0$ no solution $ii-)p|q^2-q+1$ $q|p-1 \implies p-1 = qk , p = qk+1 $ and $qk+1|q^2-q+1 \implies q^2-q+1=(qk+1)t=qkt+t$ $q^2-q+1 \equiv 1 \pmod q$ and $qkt+t \equiv t \pmod q\implies t \equiv 1 \pmod q \implies t=qx+1$ $x \ge 1 \implies qkt+t \ge q(q+1)k + q+1 \ge q(q+1)+q+1 = q^2+2q+1 > q^2-q+1 \implies $ no solution. So $x=0 \implies t = 1 \implies qkt+t = qk+1=q^2-q+1 , qk = q^2-q \implies k = q-1 \implies p=q^2-q+1$ $p,q > 3 \implies p,q \equiv 1,2 \pmod 3$ $q \equiv 2 \pmod 3 \implies q^2-q+1 \equiv p \equiv 0 \pmod 3$ so $q \equiv 1 \pmod 3 \implies q^2-q+1 \equiv 1 \equiv p \pmod 3 \implies p,q \equiv 1 \pmod 3 \implies p^3-q^5 \equiv 0 \equiv (p+q)^2 \equiv 2^2 \equiv 1 \pmod 3$ so no solution. $\implies p =3 $ or $q = 3$ $p=3 \implies q=2(p>q)$ no solution. $q=3 \implies p = 7$ So $(p,q) = (7,3)$

We claim that $(p,q)=(7,3)$ is the only possible pair. It is easy to see that $p>3$ due to size issues. Now \[ p^3-q^5=(p+q)^2 \iff p \left(p^2-p-2q \right)=q^2 \left(q^3+1 \right). \]So $p \mid q^3+1$. Thus, $q^3 \equiv -1 \pmod p$ and $q^6 \equiv 1 \pmod p$. It is now easy to show that $\text{ord}_p q=6$. Therefore, by a well-known lemma with orders and FLT, we get $6 \mid p-1$ or $p \equiv 1 \pmod 6$. The initial equation then can be manipulated to \[ 0 \equiv q^5+q^2+2q = q \left( q \left(q^3+1 \right)+2 \right) \pmod 6. \]Testing modulo $3$ yields $q=3$. Now, it is easy to conclude $p=7$.

Shortest one In the last line of my image that should be 'reach our solution'(sorry for that rip english)

This was CentroAmerican 2009 https://artofproblemsolving.com/community/c4565_2009_centroamerican

Let's suspect two cases: $ p \equiv q \pmod 3 \implies$ left side is divisible by 3, but right is not, so it is impossible. $ p \equiv -q \pmod 3 \implies$ right side is divisible by 3, but left is not, so it is also impossible. So, let $p=3$, and we have equation $(q+3)^2+q^5=27$ which has no prime solution. Let $q=3$ now, so: $p^3-(p+3)^2=243$, which is simple to solve and get $p=7$. And, finally solutions are $(p,q)=(7,3)$

$$\boxed{p^3-q^5=(p+q)^2} $$ Modulo $q$ implies $p^3\equiv p^2\mod q $ and since $\gcd (p,q)=1$ dividing by $p^2$ implies $q\mid p-1$. Now modulo $p $ implies $-q^5\equiv q^2\mod p $ this is equivalent to $q^6\equiv 1\mod p $. Notice $\gcd (q,p-1)=q $ since $q $ is prime. So we have $q^{\gcd (p-1,q)}\equiv q^q\equiv 1\mod p $. Finally we have $q^{\gcd (q,6)}\equiv 1\mod p $. Suppose $\gcd (q,6)=1$ this implies $p\mid q-1\implies q>p $ but we have $q\mid p-1$ which is $p>q $ contradiction. This means $q\in\{2,3\} $ and we can check both cases quickly and find the solution $(p,q)=(7,3) $. Note: If you haven't patient to prove cases suppose $q=2$ so this implies $q^2\equiv 1\mod p $ thus $p\mid q^2-1=(q+1)(q-1) $ so $p\mid q+1$ this implies $q+1\ge p>p-1\ge q $ so $p=q+1=3$.

Kezer wrote: We claim that $(p,q)=(7,3)$ is the only possible pair. It is easy to see that $p>3$ due to size issues. Now \[ p^3-q^5=(p+q)^2 \iff p \left(p^2-p-2q \right)=q^2 \left(q^3+1 \right). \]So $p \mid q^3+1$. Thus, $q^3 \equiv -1 \pmod p$ and $q^6 \equiv 1 \pmod p$. It is now easy to show that $\text{ord}_p q=6$. Therefore, by a well-known lemma with orders and FLT, we get $6 \mid p-1$ or $p \equiv 1 \pmod 6$. The initial equation then can be manipulated to \[ 0 \equiv q^5+q^2+2q = q \left( q \left(q^3+1 \right)+2 \right) \pmod 6. \]Testing modulo $3$ yields $q=3$. Now, it is easy to conclude $p=7$. Can you note that lemma(the statement)?

We claim that the only solution $(p,q)$ is $(7,3)$ Let $p=q$ then $p^3-p^5=(2p)^2$ but $p^3-p^5 <0$ and$(2p)^2 > 0$ which is a contradiction. So $p>q$ Using mod $3$, $p^3=0,1$mod$3$ and $q^5=0,1,2$mod$3$. Case 1:-if $p^3$=$0$ mod $3$ then $p=3$ and as $p>q$ then $q=2$ but clearly this doesn't work,so a contradiction Case 2:-$p^3$=$1$ mod $3$ $\Rightarrow$ $p=1$ mod $3$ and if $q^5=0$ mod $3$ $\Rightarrow$ $q=3$ Substituting it we get $p(p-3)(p+2)=252 \Rightarrow p=7$ Case 3:- If $p^3=1$ mod $3$ and $q=1$ mod$3$ then $q^5=1$ mod 3 $1^3-1^5=(1+1)^2$ mod$3$ which is a clear contradiction Case 4:-If $p^3=1$ mod $3$ and $q=2$ mod$3$ then $q^5=2$ mod 3 $1-2=2=(1+2)^2$ mod $3$ which is again a contradiction So we are done with the case work and our only ans is $(7,3)$

Answer:$(3,7)$ $p(p^2-p-2q)=q^2(q^3+1)$ Obviously $p>q$ so $p|q^2-q+1$ Hence $q^2(q+1)|p^2-p-2q$ $q|p(p-1) \implies q|p-1$ $p=qk+1$ We have $qk-1|q^2-q+1$ $qk-1|q^2-q+1+(qk-1)=q^2-q+qk$ $(qk-1,q)=1 \implies qk-1|q-k-1$ But $qk-1>q-k-1$ so $q=k+1$ $p=qk+1=q(q-1)+1=q^2-q+1$ $(q^2-q+1)^2=q^3+2q^2+q+1$ $q^3+q=3q^3+3$ $3|q(q^2+1)$ and $(3,q^2+1)=1$ So $q=3$ $p(p^2-p-6)=252$ If we try $p=2,3,7$ we get $p=7$.

mod 4 does not give much but 3 does

YUM! Notice that, letting $p=3$ we get $q^5+q^2+6q-18=0$ and by using IRT (https://en.wikipedia.org/wiki/Rational_root_theorem) we get no solution. Similarly, letting $q=3$ we obtain, $p^3-p^2-6p-252=0$ and IRT gives us our first solution pair $\boxed{\text (p,q)=(7,3).}$ Now, assume $p,q \equiv 1 (mod \ 3)$ then we obtain $0 \equiv 2 (mod \ 3)$ contradiction. Similarly the case $p,q \equiv (1,2);(2,1) (mod \ 3)$ gives $(2,1) \equiv 0 (mod \ 3)$ contradiction. Finally, $p,q \equiv 2 (mod \ 3)$ which gives $0 \equiv 1 (mod \ 3)$ contradiction. Thus they are our only solutions. Q.E.D $\blacksquare$

Wave-Particle wrote:

Good solution my solution is similar

The only solution is $\boxed{(7,3)}$, which works (interesting that $7^3$ and $3^5$ are the same mod $100$). Claim: One of $p,q$ is equal to $3$ Proof: Suppose not. Then both primes are not zero modulo $3$. Taking modulo $3$ gives that $p - q \equiv (p+q)^2 \pmod 3$. If $3\mid p + q$, then $3\mid p - q$, so $3\mid q$, absurd. Thus, $(p+q)^2 \equiv 1 \pmod 3$ and $p - q \equiv 1 \pmod 3$. However, this means $p \equiv 2 \pmod 3$ and $q \equiv 1 \pmod 3$, so $p + q \equiv 0 \pmod 3$, a contradiction. $\square$ Case 1: $p = 3$. Then taking modulo $q$ gives that $27 \equiv 9 \pmod q$, so $q \mid 18$, meaning $q \in \{2,3\}$. Both fail as they give $p^3 - q^5 < 0$. Case 2: $q = 3$. Then taking modulo $p$ gives that $-243 \equiv 9 \pmod p$, so $p\mid 252$. Since $p \in \{2,3\}$ give $p^3 - q^5 < 0$, $p > 3$. Thus, $p = 7$, as desired.