Answer

how did u prove it?

There are several cases , i am considering the case when $a,b>0$, Then WLOG we have $a\ge b$, now $p^2=a^3+b^3=(a+b)(a^2+b^2-ab)$, now Case :1 when $a^2+b^2-ab=1$, then we have $b^2\le a(a-b)+b^2=1$, so $b=1$ and $a(a-1)=0$, so $a=1=b$ Then $a^3+b^3=2\ne p^2$ so Case : 2 $a^2+b^2-ab$, as $a+b>1$ so we have $p|a+b$ and $p|a^2+b^2-ab=(a+b)^2-3ab\implies p|3ab$ now either $p|3\implies p=3$, and otherwise $p|a\implies p|(a+b)-a=b$, so $p^3|a^3+b^3=p^2$, contradiction, similer for $p|b$ so only case justified case is when $p=3$, then $9=3k(3^2k^2-3ab)$, where $a+b=3k$, then $1=k(3k^2-ab)$ so $k=1$ and $1=3k^2-ab$, so $ab=2, a+b=3$, so $a=2,b=1$, so $p=3$

Holmes777 wrote: how did u prove it? By factorization or by using modular arithmatic.

I don't know I'm correct or not, it is just my idea ... Letting $p$ be a prime and $a,b \in\mathbb{N}$ such that \[p^2=a^3+b^3=(a+b)(a^2-ab+b^2) \tag{1}\] If $a^2-ab+b^2=1$, then $(a-b)^2=1-ab<1$ and we easily get $a=b=1$. It is a contradiction. For $p=2$ the equation $(1)$ doesn't hold. Therefore $p\geq 3$. Hence $a$ and $b$ have different parity, say $a$ is even and $b$ is odd. From $(1)$ we obtain that \[ \begin{cases} a+b=p \\ a^2-ab+b^2=p \end{cases} \Longleftrightarrow \begin{cases} a^2+2ab+b^2=p^2 \\ a^2-ab+b^2=p \end{cases} \]Which implies that $3ab=p^2-p=p(p-1) \Longleftrightarrow p=3$ and $a=2,b=1$ Therefore the only answer is that $p=3.$ :

P=3 only. My solution is exactly like SivmengMaths.

for the second case, you can also rearrange them to get: $(a+1)(b+1) = a^2 + b^2 +1 \Rightarrow (a+1)(b+1) \geq 2ab +1$ by AM-GM. Now expand and simplify yields $(a-1)(b-1) \leq 2$, which only limits the value of a, b into a few cases, and the end you get $\boxed{3}$ as the answer

P^2 = a^3 + b^3 or p^2 =(a+b)(a^2-ab+b^2) Therefore a^2-ab +b^2 must be P or 1 since a+b cannot be P^2 But by AM GM inequality a^2+b^2 >2ab or a^2+b^-ab >2ab-ab or>ab ab>1 therefore a^2+b^2-ab>ab>1 therefore a^2 +b^2-ab =P=(a+b) assume that a>b then above equality holds if and only if b=1 and a=2 therefore (a+b)=1+2=3=P P=3(ans)

Olympus_mountaineer wrote: Holmes777 wrote: how did u prove it? By factorization or by using modular arithmatic. show us the solution

Let $p$ be the prime number and $a, b$ be the two positive integers. So we have the equation \[ p^2=a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a+b)^2-3ab) \]We can split it up into $2$ cases. Case 1: $a+b \equiv 0 \pmod{p}$ Then $a+b=p$ because if $a+b=kp$ where $k \ge 2$ its pretty easy to see that $a^3+b^3 \ge p^2$. WLOG $a \ge b$ then $a \ge \frac{p+1}{2} \implies a^3=\frac{p^3+3p^2+3p+1}{8}$. Because this must be smaller than $p^2$ we have the inequality \[ \frac{p^3+3p^2+3p+1}{8} \le p^2 \implies p^3+3p^2+3p+1 \le 8p^2 \implies p^3+3p+1 \le 5p^2 \]Now because the LHS has degree of $3$ and RHS has degree $2$ when $n$ goes to $\infty$ the LHS will be much larger than the RHS. So the only value of $p$ that satisfies this is $p=2, 3$. Testing out we only find that $p=3$ satisfies this. Case 2: $a+b \not \equiv 0 \pmod{p}$ We have $p^2=(a+b)(a^2-ab+b^2)$ so if $a+b \not \equiv 0 \pmod{p}$ then $a+b=1$ but this is a contradiction because $a, b \ge 0$ which means that $a+b>1$. So there are no solutions to this case. Our only solution is $p=3$ when $3^2=2^3+1^3$.

But what about p^2=2^3+2^3=16=4^2?

gamarjoba_naxvamdis wrote: But what about p^2=2^3+2^3=16=4^2? $p=4$ is not prime.

\[a^3 + b^3 = (a + b)(a^2 - ab + b^2) = p^2\] $p^2 = p \times p = p^2 \times 1$. But, $a + b \neq 1$ and, \[a^2 - ab + b^2 \geq 2ab - ab \geq ab \geq 1\] The equality only occurs if $(a, b) = (1, 1)$. But that set doesn't work. So, $a + b = p, a^2 - ab + b^2 = p$. \[p \mid a + b \implies p \mid a^2 + 2ab + b^2\]\[p \mid a^2 - ab + b^2 \implies p \mid 3ab\]Now, $0 < a < p$. So, $\gcd(p, a) = 1$ and same thing goes for $p$ and $b$. Hence, $\gcd(p, ab) = 1$. So, $p \mid 3$. Hence, $\boxed{p = 3}$

dwip_neel wrote: \[a^3 + b^3 = (a + b)(a^2 - ab + b^2) = p^2\] $p^2 = p \times p = p^2 \times 1$. But, $a + b \neq 1$ and, \[a^2 - ab + b^2 \geq 2ab - ab \geq ab \geq 1\] The equality only occurs if $(a, b) = (1, 1)$. But that set doesn't work. So, $a + b = p, a^2 - ab + b^2 = p$. \[p \mid a + b \implies p \mid a^2 + 2ab + b^2\]\[p \mid a^2 - ab + b^2 \implies p \mid 3ab\]Now, $0 < a < p$. So, $\gcd(p, a) = 1$ and same thing goes for $p$ and $b$. Hence, $\gcd(p, ab) = 1$. So, $p \mid 3$. Hence, $\boxed{p = 3}$ Since a+b = p we can say that a+b|p Again p|3ab so a+b|3ab => a+b|3=>a+b = 3