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Problem A prime\left trio is a collection of three prime numbers \{p, q, r\} in arithmetic progression, with common difference q-p=r-q. For example, the prime trio \{3, 5, 7\} has common difference 2.

asked Feb 8, 2023 54.8k views

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Gadam · Answer 1 · 2023-02-12T11:01:07+0000

Answer:

Proof below

Explanation:

General outline

Lemma regarding remainders of non-small primes divided by 6
Check
and
directly
Proof the rest by contradiction

Lemma

If
is prime such that
$p \\eq2$ and
$p \\eq 3$ , then
divided by 6 has remainder of 1 or 5.

Case 1: p/6 has remainder 0, 2, or 4.

Then there exists some natural number
such that either
p=6n ,
p=6n+2 , or
p=6n+4 . However,

if
, then
,
if
, then
, and
if
, then
.

By definition of divisibility,
is divisible by 2. Since
$p \\eq2$ ,
cannot be prime. This contradiction implies
cannot have a remainder of 0, 2, or 4 when divided by 6.

Case 2: p/6 has remainder 3.

Then there exists some natural number
such that
p=6n+3 . However, if
p=6n+3 , then
p=3*(2n+1) , so
is divisible by 3. Since
$p \\eq 3$ ,
is not prime. This contradiction implies
cannot have a remainder of 3 when divided by 6.

Therefore, if
is a prime number such that
$p \\eq2$ and
$p \\eq 3$ , then when divided by 6,
must have a remainder of 1 or 5.

Main proof of no prime trio with common difference of 70.

Check p=2

Consider
p=2 . Then
q=2+70=72 . However,
72=2*36 , so
is not prime. Hence,
$p \\eq2$ .

Check p=3

Consider
p=3 . Then
q=3+70=73 , and
r=73+70=143 . However,
143=11*13 , so
is not prime. Hence,
$p \\eq 3$ .

So, thus far, we've proven that if there is a prime trio with a common difference of 70, that
$p \\eq2$ and
$p \\eq 3$ .

Proof of the rest of primes by contradiction

By way of contradiction, assume that there does exist some prime trio
$\{ p,q,r \}$ such that
q-p=r-q=70 , and
$p \\eq2$ and
$p \\eq 3$ .

Then, by the Lemma proven earlier, when
is divided by 6, it must have a remainder of 1 or 5.

Case 1: p/6 has remainder 5

Assume that
has remainder 5 when divided by 6. Then, there exists some natural number
, such that
p=6n+5 .

$q=p+70\\q=(6n+5)+70\\q=6n+75\\q=3*(2n+25)$

Since
was a natural number, and the natural numbers are closed over multiplication and addition (meaning, multiplying and adding more natural numbers results in another natural number), then
is divisible by 3.

Since
is prime,
0 < p , which implies
0+70 < p+70

Since
q=p+70 ,
70 < q , so
cannot equal 3. Since
is divisible by 3, but
$q \\eq 3$ ,
is not prime, which is a contradiction to the existence of this prime trio.

Therefore, either
cannot have a remainder of 5 when divided by 6, or the prime trio does not exist.

Case 2: p/6 has remainder 1

Assume that
has remainder 1 when divided by 6. Then, there exists some natural number
, such that
p=6n+1 .

$q=p+70\\q=(6n+1)+70\\r=q+70\\r=((6n+1)+70)+70\\r=6n+141\\r=3*(2n+47)$

Since
was a natural number, and the natural numbers are closed over multiplication and addition (meaning, multiplying and adding more natural numbers results in another natural number),
is divisible by 3.

Since
is prime,
0 < q , implying
0 +70 < q+70 .

Since
r=q+70 , then
70 < r . Therefore,
$r\\eq 3$ .

Since
is divisible by 3, but
$r\\eq 3$ ,
is not prime, which is a contradiction to the existence of this prime trio (with a common difference of 70). Therefore, either
cannot have a remainder of 1 when divided by 6, or the prime trio does not exist.

Since
was prime, by our lemma
must have either a remainder of 1 or 5. Since both remainder possibilities resulted in a contradiction, our contradiction assumption is false, so there cannot exist a prime trio such that their common difference is 70.

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