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I am working on my own on the warm up problems in "Concrete Mathematics" by Knuth and others. In chapter 4, problem 9 it says to show that $(3^{77}-1)/2$ is composite. The answer key just says that $(3^{7}-1)/2$ and $(3^{11}-1)/2$ are factors but it does not explain how to find these factors. I am wondering if the number theory in that chapter could lead me to find these factors. It includes modulo arithmetic and theorems related to the number of prime factors less than a given number. How do I use number theory to find that $(3^{7}-1)/2$ is a factor of $(3^{77}-1)/2$?

Bobby Durrett
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  • See something more generic http://math.stackexchange.com/questions/262130/how-to-prove-gcdam-bm-an-bn-a-gcdm-n-b-gcdm-n – lab bhattacharjee Mar 02 '16 at 16:18
  • This seems to be a duplicate of http://math.stackexchange.com/questions/156123/show-that-frac377-12-is-odd-and-composite/156188#156188. – Dejan Govc Mar 02 '16 at 16:27

2 Answers2

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In the most elementary fashion $$\tag{eq} x^{mn} - 1 = (x^{m} - 1) (x^{m(n-1)} + x^{m(n-2)} + \dots + x^{m} + 1). $$ Take $m n = 77$ and $m = 7$, and set $x = 3$. Do the same for $11$.

In turn, you can obtain (eq) by first considering $$ y^{n} - 1 = (y - 1)(y^{n-1} + y^{n-2} + \dots + y + 1), $$ and then setting $y = x^{m}$.

Andreas Caranti
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  • Thank you. Your equation with $y^{n}-1$ is clear. It is the algebra that I didn't get in the question not the number theory taught in the chapter. – Bobby Durrett Mar 02 '16 at 16:36
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Hint $$3^{77}-1= (3^{11})^7-1^7=(3^{11}-1)(....)$$ $$3^{77}-1= (3^{7})^{11}-1^{11}=(3^{7}-1)(....)$$

Divide both sides of both equalities by 2.

N. S.
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