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Solve these problems if you're not a retard Anonymous May 31, 2024 (Fri) 19:04:43 No. 798

>>834 >>862

Solve these problems if you're not a retard:

Let

\langle a \rangle = (a_1, a_2, \ldots)

denote an infinite sequence of positive integers.

\rightarrow

Prove that there is no

\langle a \rangle

such that

\gcd(a_i + j, a_j + i) = 1

for all

i \neq j

.
Let

p \neq 2

be a prime.

\rightarrow

Prove that there is an

\langle a \rangle

such that

\gcd(a_i + j, a_j + i) = p

for all

i \neq j

Anonymous July 01, 2024 (Mon) 05:27:45 No. 834 >>860

>>798
cool homework bro

1. if the problem didnt have the +i +j bit then the solutions would just be permutations of the primes
a things position on a list is set, so we can just subtract its number to be re added later
therefore, because the naturals will always overtake the primes, any valid list must contain negative numbers

2
prime infinity
never let rules as intended into your heart

Anonymous July 11, 2024 (Thu) 15:08:06 No. 860 >>865

>>834

what the fuck

Anonymous July 12, 2024 (Fri) 12:15:42 No. 862

>>798
Okay, what now.

Anonymous July 15, 2024 (Mon) 06:40:02 No. 865

>>860
i wrote that at like 7 am
it had relative coherance at the time

Anonymous October 15, 2024 (Tue) 08:59:26 No. 999

>> 798
(A) The problems suggests that

p = 2

is the problem. If we consider

\gcd(a_{2i} + 2j, a_{2j} + 2i) = 1

, we see that at least one of

a_{2i}

a_{2j}

has to be odd — say,

a_{2i}

. Then

\gcd(a_{2i} + 2j + 1, a_{2j + 1} + 2i) = 1

implies that

a_i

is odd for odd

i

. But then

\gcd(a_{2i + 1} + 2j + 1, a_{2j + 1} + 2j + 1) = 1

is a contradiction since both sides are divisible by

2

.
(B) Just let

a_i = pi - i + 1 > 0

, which gives

\gcd(pi - i + 1 + j, pj - j + 1 + i) = \gcd(p(i + j) + 2, pj - j + 1 + i)

and

p \nmid p(i + j) + 2

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