HCF of (2^125-1)(2^55-1) is?
Answers
Answer:
The polynomials x100−1 and x120−1 are both divisible by x20−1, but have no other polynomial factors in common. So we can tell right away that 220−1 is a divisor of both numbers. But are there any larger divisors? If there’s a deep number-theoretical reason why there can’t be any larger divisors, I’ll be interested to know! In the meantime, I’ll plug in x=2, and prime-factorize the heck out of this thing.
If we let y=x20, then x100−1=y5–1=(y−1)(y4+y3+y2+y+1).
Letting y=220, we’ll check the prime factorizations of the polynomial factors of y5–1y−1 and y6–1y−1 have any prime factors in common.
The prime factorization of y5–1y−1 is 5 × 101 × 251 × 601 × 1801 × 4051 × 8101 × 268501
And x120−1=y6–1=(y−1)(y+1)(y2−y+1)(y2+y+1).
Still letting y=220, the prime factorizations of the factors of y6–1y−1 are
(y+1)=17×61681
(y2−y+1)=241×4562284561
(y2+y+1)=3×7×13×61×151×331×1321
Whew! That was hard. (Not really, Wolfram Alpha did the heavy lifting.) So we see there are no prime factors in common of y5–1y−1 and y6–1y−1 when we let y=220.
- So the answer to the question is 220−1, or 1048575.