prove that there is no Arithmetic progression which consist only of prime number.?

Question

don't prove by observation. use properties of prime numbers and arithmetic progression.
prove by contradiction if possible.

Answer 1

n th term of AP is a +(n-1) d

Consider the case a is positive integer


Then when n = a+1

a will be a factor of a +(n-1) d & hence it will not be a prime

Then consider a is negetive interger then also when (n-1) becomes -a , it will have a factor

a & d can not be fractions since all terms must be whole numbers


Hence no set of prime numbers will fall on an AP series

Examples

3,5,7,(9) ..... It terminated with 9
3,7,11,(15) .............................15
3,11,19,(27)...............................27
3,13,23,(33)...............................33
5,11,17,23,29,(35).....................35

Answer 2

Assume a, a + r, a + 2r, a + 3r... is an arithmetic progression with all prime numbers. Then "a" has to be prime, which means it's an integer. Similarly, r has to be an integer, in order for both a and a+r to be integers (let alone prime).

Since the coefficients in front of r follow the series 1, 2, 3, ..., then eventually in the series, you'll find the term a + a*r. This is divisible by a, and thus isn't prime. So no arithmetic progression can be a set of all prime numbers, because eventually there will be a term that's a multiple of the first term.

Answer 3

Proof by contradiction:

The general form of an Arithmetic Progression, where the first term is a and the common diference is d

a, a + d, a + 2d.

List the first few primes:

2, 3, 5, 7, 11

Here, a = 2
d = 3 - 2 = 1

But the difference between the third and second term is 5 - 3 = 2

Which contradicts the fact that d = 1, which was derived from the first two terms.

We get this contradiction because what we assumed is wrong. Hence, No Arithmetic Progression can consist only of Prime numbers.

Answer 4

Proof by contradiction.
Assume such a progression exists. It can be described as:

a, a+n, a+2n, a+3n, ... where n is the common difference and all the terms are prime
At some point, the term a + an will arise.
a+an = a(1+n) and is therefore not prime.
Contradiction.

NOTE that n≠0 is an unmentioned premise of the problem or else (degenerate) progressions do exist!
2, 2, 2, 2, 2, .... for example.
Also, n<0 is impossible as terms would become negative eventually and there are no 'negative' primes.

ksoileau I disagree. your "touchups" are superfluous.
They are either unnecessary for the proof to work or implied by the problem or definitions of terms.

Answer 5

Zaphod is nearly right, just needs a slight touchup to the proof:

Proof by contradiction.
Assume such a progression exists. It can be described as:
a, a+n, a+2n, a+3n, ... where
* n>0 *
is the common difference and all the terms are prime
* and a>1 since a is prime. *
At some point, the term a + an will arise.
a+an = a(1+n)
* and since a>1 and 1+n>1, *
a+an is therefore not prime.
Contradiction.

Answer 6

Consider a series
a1, a1 + d, a1 + 2d etc etc

In order for all these terms to be prime,
1) d must be even
2) an must all be odd
3) a1 must be a positive odd integer >= 3

So the n+1 th term is a1 + n*d
since n = 1,2,3,
there must be a value of n, k such that k = a1
and thus a_k+1 = a1 + a1*d = a1*(1+d)
So this particular term is not prime.