Question

25. Ifn e N and 2n - 1 is prime, then n is prime. Proof. Assume n is not prime. Write n-ab for some a,b >1. Then 2n-1- 2b-1-(2b-1) (2ab-b +2ab-2b +2ab-3b +...+2ab-ab). Hence 2"-1 s composite. . 29. If integers a and b are not both zero, then gcd(a,b)-gcd(a - b,b) Proof (Direct) Suppose integers α and b are not both zero. Let d=gcd(a, b). Because d is a divisor of both a and b, we have a - dx and b -dy for some integers x and y. Then a-b-dx-dy=d(x-y), so it follows that d is also a common divisor of a -b and b. Therefore it can't be greater than the greatest common divisor of a - b and b, which is to say gcd(a,b)- d s gcd(a - b,b) Now let e = gcd(a-b,b). Then e divides both a-b and b, that is, a-b = ex and b-ey for integers x and y. Then α-(a-b)+b-ex + ey-e(x+y), so now we see that e is a divisor of both α and b. Thus it is not more than their greatest common divisor, that is, gcd(a - b,b) e s gcd(a,b). The above two paragraphs have given gcd(a,b)s gcd(a - b,b) and gcd(a - b,b)s ged(a,b). Thus gcd(a, b) - gcd(a - b,b)

For number 25, can someone explain to me how they got (2^(ab-b)+2^(ab-2b)+2^(ab-3b)+...+(2^(ab-ab)) and how they reached to that conclusion?

For number 29, can someone explain to me how "it can't be greater than the greatest common divisor of a-b and b"? I would think that gcd(a, b) would be greater than gcd(a-b, b) because "a" and "b" are bigger than "a-b" and "b" so that confused me. Thank you!

Answer 1

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