TM2 LEVEL A 2024 P4) Find all the triples (a, b, c) of positive integers such that a³ + b³+ c³ = 2p, p prime, and 3ab = 2c². Notice that 3ab = 2c² implies 3abc = 2c³, and therefore a³ + b³ + c³= a³ + b³ + (-c)³ - (3ab(-c)) = (a + b - c)(a² + b² + c² - ab + ac + bc) = 2p. And (a² + b² + c² -ab + ac + bc) >= ((a - b)²+ (b - c)² + (c - a)²)/2 + 2ac + 2bc > a + b > a + b - c. Case 1. a + b = c + 2 a² + b² + c² - ab + ac + bc = p. 3ab = 2c² => (a + b)² - 3ab + c² + c(a + b) = p => (c + 2)² - 2c² + c²+ c(c + 2) = p => c² + 6c + 4 = p => 2c² + 12c + 8 = 2p = a³ + b³ + c³ > c³ => 2c² + 12c + 8 > c³. Notice that 3ab = 2c² => c is divisible by 3, and c >= 6 does not satisfy the inequality => c = 3 => p = 31. 3ab = 2c² = 18 => ab = 6, a³ + b³ + c³ = 2p => a³ + b³ = 62 - 27 = 35 => a = 2, b = 3 or a = 3, b = 2. (2, 3, 3); (3, 2, 2) are solutions. Case 2. a + b = c + 1 a² + b² + c² - ab + bc + ca = 2p => (a + b)² - 3ab + c² + c(a + b) = 2p => (c + 1)² - 2c² + c² + c(c + 1) = 2p => c² + 2c + 1 - 2c² + c² + c² + c = 2p => c² + 3c + 1 = 2p = a³ + b³ + c³ > c³ => Not even c = 3 satisfies the inequality. So the only solutions are (2, 3, 3) and (3, 2, 3).

Similar to @above, but anyways: $a^3+b^3+c^3=a^3+b^3-c^3+2c^3=a^3+b^3-c^3+3abc=\boxed{(a+b-c)(a^2+b^2+c^2-ab+bc+ca)=2p}$. Notice that $(a+b-c)\equiv a^3+b^3+c^3\equiv 0 \ (mod \ 2)$, and $(a^2+b^2+c^2-ab+bc+ca)> 1 \ $ trivially, thus $\boxed{c=a+b-2}$. (Clearly $p$ is odd.) $3ab=2c^2\Rightarrow 3ab=2(a+b-2)^2\geq 2(2\sqrt{ab}-2)^2\Rightarrow 5(\sqrt{ab})^2-16\sqrt{ab}+8\leq 0\implies \sqrt{ab}\leq\frac{16+\sqrt{96}}{10}<3\Rightarrow \boxed{ab\leq 8}$. But simple manipulation of $3ab=2c^2$ yields $6\mid ab\implies \boxed{ab=6}$, which gives us $(a, b, c)=\{(2, 3, 3); (3, 2, 3)\} \ _{\blacksquare}$