Exploring the Primality of 3^(2520)+4^(4038)

[anemone]

Is $3^{2520}+4^{4038}$ a prime?

 

[kaliprasad]

Spoiler

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]

Spoiler

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

 

[Greg]

Would you mind telling me what theorem this is based on?

Spoiler

Sum of cubes.

\(\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2)\)

Proof:

\(\displaystyle (a+b)^3=a^3+3a^2b+3ab^2+b^3\)

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

 

[topsquark]

Spoiler

Sum of cubes.

\(\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2)\)

Proof:

\(\displaystyle (a+b)^3=a^3+3a^2b+3ab^2+b^3\)

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

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

 

[anemone]

Spoiler

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)