The discussion revolves around the primality of the expression $3^{2520}+4^{4038}$. Participants explore the underlying theories and theorems related to this mathematical inquiry.
The discussion does not reach a consensus on the primality of the expression, and multiple viewpoints regarding the underlying theories and interpretations remain present.
Some assumptions about theorems and interpretations of previous statements are not fully articulated, which may affect the clarity of the discussion.
Would you mind telling me what theorem this is based on?kaliprasad said:we know $GCD(2520,4038)=6$ and 3 is odd factor of 6 $2520 = 3 * 840, 4038 = 3* 1346$
Hence $3^{2520} + 4^{4038}= (3^{840})^3 + (4^{1346})^3= (3^{840} + 4^{1346})((3^{840})^2 - 3 ^{840} * 4 ^ {1346} + (4^{1346})^2)$
2 factors both $> 1$ hence not a prime
topsquark said:Would you mind telling me what theorem this is based on?
Sorry, I should have been more specific. I couldn't see how the last line and the line above it solved the problem. I do now...I was misinterpreting the last line.greg1313 said:Sum of cubes.
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$
Proof:
$$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
$$\begin{align*}a^3+b^3&=(a+b)^3-3a^2b+3ab^2 \\
&=(a+b)^3-3ab(a+b) \\
&=(a+b)((a+b)^2-3ab) \\
&=(a+b)(a^2-ab+b^2)\end{align*}$$
kaliprasad said:we know $GCD(2520,4038)=6$ and 3 is odd factor of 6 $2520 = 3 * 840, 4038 = 3* 1346$
Hence $3^{2520} + 4^{4038}= (3^{840})^3 + (4^{1346})^3= (3^{840} + 4^{1346})((3^{840})^2 - 3 ^{840} * 4 ^ {1346} + (4^{1346})^2)$
2 factors both $> 1$ hence not a prime