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Answer:
(a) Let us recall the division algorithm: given two positive integers and
there exist other two positive integers
and
such that
where
and
is called the <em>remainder</em>.
So, given any positive integer and 3 we can write
where
. Thus, every
can be written as
Now, notice that . Hence, every number can be written as
, or
or
.
A number is prime if and only if its only factors are 1 and
itself. So, a number of the form
cannot be prime. Therefore, every primer number is of the form
or
.
(b) Assume that there are three prime numbers such that ,
and
are prime.
By the previous exercise or
. Let us analyze both cases separately.
<em>First case</em>: . Then
that can be prime, and
that is not prime. Hence, there are not such three primes with
.
<em>Second case</em>: . Then,
that can be prime, and
that cannot be prime. Hence, there are not such three primes with
.
Therefore, there are no three primes of the form ,
and
, except for 3, 5 and 7.
Notice that this is only possible because 5=2*3-1 and 2*3-3=3, that is the only ‘‘multiple’’ of 3 that is prime.
Answer:
Table B
Step-by-step explanation:
The table represents a linear function if the ratio of change in y (∆y) to change in x (∆x) is a constant.
A — first two points: ∆y/∆x = (1-2)/(3-0) = -1/3
second two points: ∆y/∆x = (6-1)/(4-3) = 5 ≠ -1/3
__
B — first two points: ∆y/∆x = (2-(-3))/(4-(-1)) = 5/5 = 1
second two points: ∆y/∆x = (4-2)/(6-4) = 2/2 = 1, the same as for the first points. This is the table that answers the question.
__
C — first two points: ∆y/∆x = (0-(-2))/(0-(-3)) = 2/3
second two points: ∆y/∆x = (4-0)/(2-0) = 4/2 = 2 ≠ 2/3
__
D — first two points: ∆y/∆x = (-2-(-7))/(0-5) = 5/-5 = -1
second two points: ∆y/∆x = (2-(-2))/(2-0) = 4/2 = 2 ≠ -1
