MHB Is $3^{2520}+4^{4038}$ Prime?

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The discussion centers on whether the expression $3^{2520}+4^{4038}$ is prime. Participants seek clarification on the theorem applicable to this problem. Initial confusion about the solution's explanation is acknowledged, but understanding is eventually achieved. One user praises another for their explanation. The conversation highlights the importance of mathematical clarity in determining the primality of complex expressions.
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Is $3^{2520}+4^{4038}$ a prime?
 
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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
 
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
Would you mind telling me what theorem this is based on?

Thanks!

-Dan
 
topsquark said:
Would you mind telling me what theorem this is based on?

Sum of cubes.

$$a^3+b^3=(a+b)(a^2-ab+b^2)$$

Proof:

$$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$

$$\begin{align*}a^3+b^3&=(a+b)^3-3a^2b+3ab^2 \\
&=(a+b)^3-3ab(a+b) \\
&=(a+b)((a+b)^2-3ab) \\
&=(a+b)(a^2-ab+b^2)\end{align*}$$
 
greg1313 said:
Sum of cubes.

$$a^3+b^3=(a+b)(a^2-ab+b^2)$$

Proof:

$$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$

$$\begin{align*}a^3+b^3&=(a+b)^3-3a^2b+3ab^2 \\
&=(a+b)^3-3ab(a+b) \\
&=(a+b)((a+b)^2-3ab) \\
&=(a+b)(a^2-ab+b^2)\end{align*}$$
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.

Thanks!

-Dan
 
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

Very well done, kaliprasad!(Cool)
 
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