Prime Factorization of 49 + 39 - MathFest 2004

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SUMMARY

The prime factorization of the expression 49 + 39 can be efficiently obtained using the sum of cubes formula. The expression can be rewritten as 4^9 + 3^9, which simplifies to (4^3)^3 + (3^3)^3. This leads to the factors (91) and (64² - 64·27 + 27²). The complete factorization results in 7, 13, 19, and 163 after applying trial division on the factors derived from the sum of cubes. This method provides a systematic approach to factorization without excessive multiplication.

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  • Knowledge of trial division for breaking down factors
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devious_
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Is there a method one can use to obtain the prime factorization of a certain number?

For example:
Find the prime factorization of 49 + 39. [MathFest 2004]

I realize that I can re-write the expression as 29.29+39, but that's about as far as I can go. :cry:
 
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For this specific number, notice it's a sum of cubes, [tex]4^9+3^9=(4^3)^3+(3^3)^3[/tex] and use the general formula for the sum of cubes. This breaks it into 2 factors nicely, both are much easier to deal with then the orignial. Just use trial division on these 2 factors to furthur break them down.
 
I did use the sum of two cubes formula. This is what I got:
64³+27³=(91)(64²-64.27+27²)=(7)(13)(4096-1728+729)=7.13.19.163

That's the correct answer, but I had to multiply the second bracket out to get it. I was just wondering if there was some other way I could use that wouldn't involve this level of multiplication.
 

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