Showing that Goldbach's Conjecture is correct in certain casesSum of odd prime and odd semiprime as sum of two odd primes?Property of a sequence involving near-primesquestion about prime numbersA conjecture about prime numbersTo find $beta_1, beta_2$ in order to satisfy $2^2 p_1^a_1+1-beta_1p_2^a_2+1-beta_2p_3^a_3p_4^a_4=(p_1-1)(p_2-1)$Related to proving Goldbach's ConjectureGenerative function for Goldbach's conjectureGoldbach's Conjecture and 1-1 correspondenceHow many pairs of adjacent numbers each have a prime number of factors?Goldbach's conjecture and Diophantine equations.
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Showing that Goldbach's Conjecture is correct in certain cases
Sum of odd prime and odd semiprime as sum of two odd primes?Property of a sequence involving near-primesquestion about prime numbersA conjecture about prime numbersTo find $beta_1, beta_2$ in order to satisfy $2^2 p_1^a_1+1-beta_1p_2^a_2+1-beta_2p_3^a_3p_4^a_4=(p_1-1)(p_2-1)$Related to proving Goldbach's ConjectureGenerative function for Goldbach's conjectureGoldbach's Conjecture and 1-1 correspondenceHow many pairs of adjacent numbers each have a prime number of factors?Goldbach's conjecture and Diophantine equations.
$begingroup$
I try to prove the following formula (this is a formula that proves Goldbach's Conjecture in a limited number of cases): $$forall x in mathbb N , exists (y in mathbb N , p_1) : 2^xy – p_1 = p_2$$ when $x in mathbb N$ is a prime number.
I've written a complete proof, but I'm not sure if my proof is correct or not.
Any help will be highly appreciated!
The Proof
Let us prove the following formula: $$forall x in mathbb N , exists (y in mathbb N , p_1) : 2^xy – p_1 = p_2$$ Given that x is a prime number, y is a natural number, and $p_1,p_2$ are primes.
Let us suppose that the opposite formula is correct: $exists x in mathbb N , forall ( y in mathbb N , p_1) : 2^xy – p_1 ne p_2$
$2^xy ne p_1 + p_2 = 2m$
The numbers $p_1$ and $p_2$ are independent of each other (because we have an inequality). For this reason, $2m$ represents all the even numbers for which Goldbach’s conjecture is correct. That is: $4 ≤ 2m ≤ 4 ∗ 10^18$
Suppose that $x$ is a prime. Let us choose $p_1 = x$ and $y = 1$. It follows that:
$2^p_1 ne p_1 + p_2 implies 2^p_1 = p_1 + 2m + 1 implies 2^p_1 – 1 = p_1 + 2m$
Since $p_1$ is a prime number, all the prime factors of $2^p_1 – 1$ are larger than $p_1$. It follows that:
$p_3∗ p_4∗ ... ∗ p_3+k = 2m + p_1 | p_3,p_4,... ,p_3+k > p_1 , k ∈ mathbb N$
$2m = p_3∗ p_4∗ ... ∗ p_3+k – p_1 > p_1^k – p_1$
$2m > p_1^k – p_1 = p_1(p_1^k-1 - 1)$
From step 3 and step 9, we know that $p_1(p_1^k-1 - 1) < 4$, hence:
Either $p_1 < 2$ or $p_1^k-1 – 1 < 2$
If $p_1 < 2$, we have a contradiction.
If $p_1^k-1 – 1 < 2$ then $p_1^k-1 < 3$, hence $p_1^k-1 = 2$. It means that $p_1 = 2$ and $k = 2$.
The number $2^p_1 – 1$ is odd. Therefore, its prime factors $(p_3,p_4,... ,p_3+k)$ are odd as well.
Steps 7,13,14 result in a contradiction, because the equality $p_3∗ p_4∗ ... ∗ p_3+k = 2m + p_1$ means that $[Odd number] = [Even number]$.
Because of the contradictions that we have in step 15 and step 12, the supposition in step 1 is incorrect. Hence, the original formula is correct.
Q.E.D
number-theory proof-verification
asked Mar 13 at 18:55
Matan CohenMatan Cohen
14
$endgroup$
|
show 1 more comment
$begingroup$
I try to prove the following formula (this is a formula that proves Goldbach's Conjecture in a limited number of cases): $$forall x in mathbb N , exists (y in mathbb N , p_1) : 2^xy – p_1 = p_2$$ when $x in mathbb N$ is a prime number.
I've written a complete proof, but I'm not sure if my proof is correct or not.
Any help will be highly appreciated!
The Proof
Let us prove the following formula: $$forall x in mathbb N , exists (y in mathbb N , p_1) : 2^xy – p_1 = p_2$$ Given that x is a prime number, y is a natural number, and $p_1,p_2$ are primes.
Let us suppose that the opposite formula is correct: $exists x in mathbb N , forall ( y in mathbb N , p_1) : 2^xy – p_1 ne p_2$
$2^xy ne p_1 + p_2 = 2m$
The numbers $p_1$ and $p_2$ are independent of each other (because we have an inequality). For this reason, $2m$ represents all the even numbers for which Goldbach’s conjecture is correct. That is: $4 ≤ 2m ≤ 4 ∗ 10^18$
Suppose that $x$ is a prime. Let us choose $p_1 = x$ and $y = 1$. It follows that:
$2^p_1 ne p_1 + p_2 implies 2^p_1 = p_1 + 2m + 1 implies 2^p_1 – 1 = p_1 + 2m$
Since $p_1$ is a prime number, all the prime factors of $2^p_1 – 1$ are larger than $p_1$. It follows that:
$p_3∗ p_4∗ ... ∗ p_3+k = 2m + p_1 | p_3,p_4,... ,p_3+k > p_1 , k ∈ mathbb N$
$2m = p_3∗ p_4∗ ... ∗ p_3+k – p_1 > p_1^k – p_1$
$2m > p_1^k – p_1 = p_1(p_1^k-1 - 1)$
From step 3 and step 9, we know that $p_1(p_1^k-1 - 1) < 4$, hence:
Either $p_1 < 2$ or $p_1^k-1 – 1 < 2$
If $p_1 < 2$, we have a contradiction.
If $p_1^k-1 – 1 < 2$ then $p_1^k-1 < 3$, hence $p_1^k-1 = 2$. It means that $p_1 = 2$ and $k = 2$.
The number $2^p_1 – 1$ is odd. Therefore, its prime factors $(p_3,p_4,... ,p_3+k)$ are odd as well.
Steps 7,13,14 result in a contradiction, because the equality $p_3∗ p_4∗ ... ∗ p_3+k = 2m + p_1$ means that $[Odd number] = [Even number]$.
Because of the contradictions that we have in step 15 and step 12, the supposition in step 1 is incorrect. Hence, the original formula is correct.
Q.E.D
number-theory proof-verification
asked Mar 13 at 18:55
Matan CohenMatan Cohen
14
$endgroup$
1
$begingroup$
Your step 10 doesn't make any sense. If Goldbach is true for $2m le 4* 10^18$ then it is true for $2^n le 4*10^18$, that's what you have shown
$endgroup$
– reuns
Mar 13 at 19:39
$begingroup$
2^n can be greater than 2m
$endgroup$
– Matan Cohen
Mar 13 at 20:37
$begingroup$
The "proof" is a long chain of obfuscation that proves that $2^p$ is a sum of two primes when $2^p$ is small enough so we know that by computation
$endgroup$
– Conrad
Mar 13 at 22:10
$begingroup$
Conrad, I'm afraid you didn't understand my proof at all. I assumed that I can prove Goldbach for small numbers only, up to 2m (see step 2). But I got a contradiction in step 15, this means that 2^(xy) = p1 + p2. Since the size of x is unlimited by definition, 2^(xy) can be as large as you want, much more than any computer can compute...
$endgroup$
– Matan Cohen
Mar 13 at 22:55
1
$begingroup$
Step 3 is utter nonsense as written (all the even numbers for which Goldbach is true being less than whatever) since there are tons of even numbers bigger than 4*10^18 that are sums of two primes, like say $p+3$, $p$ arbitrary prime going to infinity; this is not even the type of subtle fallacy that appears in the RH case and requires thought to figure out, but complete misunderstanding of elementary math.
$endgroup$
– Conrad
Mar 15 at 1:51
|
show 1 more comment
$begingroup$
I try to prove the following formula (this is a formula that proves Goldbach's Conjecture in a limited number of cases): $$forall x in mathbb N , exists (y in mathbb N , p_1) : 2^xy – p_1 = p_2$$ when $x in mathbb N$ is a prime number.
I've written a complete proof, but I'm not sure if my proof is correct or not.
Any help will be highly appreciated!
The Proof
Let us prove the following formula: $$forall x in mathbb N , exists (y in mathbb N , p_1) : 2^xy – p_1 = p_2$$ Given that x is a prime number, y is a natural number, and $p_1,p_2$ are primes.
Let us suppose that the opposite formula is correct: $exists x in mathbb N , forall ( y in mathbb N , p_1) : 2^xy – p_1 ne p_2$
$2^xy ne p_1 + p_2 = 2m$
The numbers $p_1$ and $p_2$ are independent of each other (because we have an inequality). For this reason, $2m$ represents all the even numbers for which Goldbach’s conjecture is correct. That is: $4 ≤ 2m ≤ 4 ∗ 10^18$
Suppose that $x$ is a prime. Let us choose $p_1 = x$ and $y = 1$. It follows that:
$2^p_1 ne p_1 + p_2 implies 2^p_1 = p_1 + 2m + 1 implies 2^p_1 – 1 = p_1 + 2m$
Since $p_1$ is a prime number, all the prime factors of $2^p_1 – 1$ are larger than $p_1$. It follows that:
$p_3∗ p_4∗ ... ∗ p_3+k = 2m + p_1 | p_3,p_4,... ,p_3+k > p_1 , k ∈ mathbb N$
$2m = p_3∗ p_4∗ ... ∗ p_3+k – p_1 > p_1^k – p_1$
$2m > p_1^k – p_1 = p_1(p_1^k-1 - 1)$
From step 3 and step 9, we know that $p_1(p_1^k-1 - 1) < 4$, hence:
Either $p_1 < 2$ or $p_1^k-1 – 1 < 2$
If $p_1 < 2$, we have a contradiction.
If $p_1^k-1 – 1 < 2$ then $p_1^k-1 < 3$, hence $p_1^k-1 = 2$. It means that $p_1 = 2$ and $k = 2$.
The number $2^p_1 – 1$ is odd. Therefore, its prime factors $(p_3,p_4,... ,p_3+k)$ are odd as well.
Steps 7,13,14 result in a contradiction, because the equality $p_3∗ p_4∗ ... ∗ p_3+k = 2m + p_1$ means that $[Odd number] = [Even number]$.
Because of the contradictions that we have in step 15 and step 12, the supposition in step 1 is incorrect. Hence, the original formula is correct.
Q.E.D
number-theory proof-verification
asked Mar 13 at 18:55
Matan CohenMatan Cohen
14
$endgroup$
I try to prove the following formula (this is a formula that proves Goldbach's Conjecture in a limited number of cases): $$forall x in mathbb N , exists (y in mathbb N , p_1) : 2^xy – p_1 = p_2$$ when $x in mathbb N$ is a prime number.
I've written a complete proof, but I'm not sure if my proof is correct or not.
Any help will be highly appreciated!
The Proof
Let us prove the following formula: $$forall x in mathbb N , exists (y in mathbb N , p_1) : 2^xy – p_1 = p_2$$ Given that x is a prime number, y is a natural number, and $p_1,p_2$ are primes.
Let us suppose that the opposite formula is correct: $exists x in mathbb N , forall ( y in mathbb N , p_1) : 2^xy – p_1 ne p_2$
$2^xy ne p_1 + p_2 = 2m$
The numbers $p_1$ and $p_2$ are independent of each other (because we have an inequality). For this reason, $2m$ represents all the even numbers for which Goldbach’s conjecture is correct. That is: $4 ≤ 2m ≤ 4 ∗ 10^18$
Suppose that $x$ is a prime. Let us choose $p_1 = x$ and $y = 1$. It follows that:
$2^p_1 ne p_1 + p_2 implies 2^p_1 = p_1 + 2m + 1 implies 2^p_1 – 1 = p_1 + 2m$
Since $p_1$ is a prime number, all the prime factors of $2^p_1 – 1$ are larger than $p_1$. It follows that:
$p_3∗ p_4∗ ... ∗ p_3+k = 2m + p_1 | p_3,p_4,... ,p_3+k > p_1 , k ∈ mathbb N$
$2m = p_3∗ p_4∗ ... ∗ p_3+k – p_1 > p_1^k – p_1$
$2m > p_1^k – p_1 = p_1(p_1^k-1 - 1)$
From step 3 and step 9, we know that $p_1(p_1^k-1 - 1) < 4$, hence:
Either $p_1 < 2$ or $p_1^k-1 – 1 < 2$
If $p_1 < 2$, we have a contradiction.
If $p_1^k-1 – 1 < 2$ then $p_1^k-1 < 3$, hence $p_1^k-1 = 2$. It means that $p_1 = 2$ and $k = 2$.
The number $2^p_1 – 1$ is odd. Therefore, its prime factors $(p_3,p_4,... ,p_3+k)$ are odd as well.
Steps 7,13,14 result in a contradiction, because the equality $p_3∗ p_4∗ ... ∗ p_3+k = 2m + p_1$ means that $[Odd number] = [Even number]$.
Because of the contradictions that we have in step 15 and step 12, the supposition in step 1 is incorrect. Hence, the original formula is correct.
Q.E.D
number-theory proof-verification
number-theory proof-verification
asked Mar 13 at 18:55
Matan CohenMatan Cohen
14
asked Mar 13 at 18:55
Matan CohenMatan Cohen
14
asked Mar 13 at 18:55
Matan CohenMatan Cohen
14
asked Mar 13 at 18:55
Matan CohenMatan Cohen
14
asked Mar 13 at 18:55
Matan CohenMatan Cohen
14
14
1
$begingroup$
Your step 10 doesn't make any sense. If Goldbach is true for $2m le 4* 10^18$ then it is true for $2^n le 4*10^18$, that's what you have shown
$endgroup$
– reuns
Mar 13 at 19:39
$begingroup$
2^n can be greater than 2m
$endgroup$
– Matan Cohen
Mar 13 at 20:37
$begingroup$
The "proof" is a long chain of obfuscation that proves that $2^p$ is a sum of two primes when $2^p$ is small enough so we know that by computation
$endgroup$
– Conrad
Mar 13 at 22:10
$begingroup$
Conrad, I'm afraid you didn't understand my proof at all. I assumed that I can prove Goldbach for small numbers only, up to 2m (see step 2). But I got a contradiction in step 15, this means that 2^(xy) = p1 + p2. Since the size of x is unlimited by definition, 2^(xy) can be as large as you want, much more than any computer can compute...
$endgroup$
– Matan Cohen
Mar 13 at 22:55
1
$begingroup$
Step 3 is utter nonsense as written (all the even numbers for which Goldbach is true being less than whatever) since there are tons of even numbers bigger than 4*10^18 that are sums of two primes, like say $p+3$, $p$ arbitrary prime going to infinity; this is not even the type of subtle fallacy that appears in the RH case and requires thought to figure out, but complete misunderstanding of elementary math.
$endgroup$
– Conrad
Mar 15 at 1:51
|
show 1 more comment
1
$begingroup$
Your step 10 doesn't make any sense. If Goldbach is true for $2m le 4* 10^18$ then it is true for $2^n le 4*10^18$, that's what you have shown
$endgroup$
– reuns
Mar 13 at 19:39
$begingroup$
2^n can be greater than 2m
$endgroup$
– Matan Cohen
Mar 13 at 20:37
$begingroup$
The "proof" is a long chain of obfuscation that proves that $2^p$ is a sum of two primes when $2^p$ is small enough so we know that by computation
$endgroup$
– Conrad
Mar 13 at 22:10
$begingroup$
Conrad, I'm afraid you didn't understand my proof at all. I assumed that I can prove Goldbach for small numbers only, up to 2m (see step 2). But I got a contradiction in step 15, this means that 2^(xy) = p1 + p2. Since the size of x is unlimited by definition, 2^(xy) can be as large as you want, much more than any computer can compute...
$endgroup$
– Matan Cohen
Mar 13 at 22:55
1
$begingroup$
Step 3 is utter nonsense as written (all the even numbers for which Goldbach is true being less than whatever) since there are tons of even numbers bigger than 4*10^18 that are sums of two primes, like say $p+3$, $p$ arbitrary prime going to infinity; this is not even the type of subtle fallacy that appears in the RH case and requires thought to figure out, but complete misunderstanding of elementary math.
$endgroup$
– Conrad
Mar 15 at 1:51
1
1
$begingroup$
Your step 10 doesn't make any sense. If Goldbach is true for $2m le 4* 10^18$ then it is true for $2^n le 4*10^18$, that's what you have shown
$endgroup$
– reuns
Mar 13 at 19:39
$begingroup$
Your step 10 doesn't make any sense. If Goldbach is true for $2m le 4* 10^18$ then it is true for $2^n le 4*10^18$, that's what you have shown
$endgroup$
– reuns
Mar 13 at 19:39
$begingroup$
2^n can be greater than 2m
$endgroup$
– Matan Cohen
Mar 13 at 20:37
$begingroup$
2^n can be greater than 2m
$endgroup$
– Matan Cohen
Mar 13 at 20:37
$begingroup$
The "proof" is a long chain of obfuscation that proves that $2^p$ is a sum of two primes when $2^p$ is small enough so we know that by computation
$endgroup$
– Conrad
Mar 13 at 22:10
$begingroup$
The "proof" is a long chain of obfuscation that proves that $2^p$ is a sum of two primes when $2^p$ is small enough so we know that by computation
$endgroup$
– Conrad
Mar 13 at 22:10
$begingroup$
Conrad, I'm afraid you didn't understand my proof at all. I assumed that I can prove Goldbach for small numbers only, up to 2m (see step 2). But I got a contradiction in step 15, this means that 2^(xy) = p1 + p2. Since the size of x is unlimited by definition, 2^(xy) can be as large as you want, much more than any computer can compute...
$endgroup$
– Matan Cohen
Mar 13 at 22:55
$begingroup$
Conrad, I'm afraid you didn't understand my proof at all. I assumed that I can prove Goldbach for small numbers only, up to 2m (see step 2). But I got a contradiction in step 15, this means that 2^(xy) = p1 + p2. Since the size of x is unlimited by definition, 2^(xy) can be as large as you want, much more than any computer can compute...
$endgroup$
– Matan Cohen
Mar 13 at 22:55
1
1
$begingroup$
Step 3 is utter nonsense as written (all the even numbers for which Goldbach is true being less than whatever) since there are tons of even numbers bigger than 4*10^18 that are sums of two primes, like say $p+3$, $p$ arbitrary prime going to infinity; this is not even the type of subtle fallacy that appears in the RH case and requires thought to figure out, but complete misunderstanding of elementary math.
$endgroup$
– Conrad
Mar 15 at 1:51
$begingroup$
Step 3 is utter nonsense as written (all the even numbers for which Goldbach is true being less than whatever) since there are tons of even numbers bigger than 4*10^18 that are sums of two primes, like say $p+3$, $p$ arbitrary prime going to infinity; this is not even the type of subtle fallacy that appears in the RH case and requires thought to figure out, but complete misunderstanding of elementary math.
$endgroup$
– Conrad
Mar 15 at 1:51
|
show 1 more comment
0
active
oldest
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1
$begingroup$
Your step 10 doesn't make any sense. If Goldbach is true for $2m le 4* 10^18$ then it is true for $2^n le 4*10^18$, that's what you have shown
$endgroup$
– reuns
Mar 13 at 19:39
$begingroup$
2^n can be greater than 2m
$endgroup$
– Matan Cohen
Mar 13 at 20:37
$begingroup$
The "proof" is a long chain of obfuscation that proves that $2^p$ is a sum of two primes when $2^p$ is small enough so we know that by computation
$endgroup$
– Conrad
Mar 13 at 22:10
$begingroup$
Conrad, I'm afraid you didn't understand my proof at all. I assumed that I can prove Goldbach for small numbers only, up to 2m (see step 2). But I got a contradiction in step 15, this means that 2^(xy) = p1 + p2. Since the size of x is unlimited by definition, 2^(xy) can be as large as you want, much more than any computer can compute...
$endgroup$
– Matan Cohen
Mar 13 at 22:55
1
$begingroup$
Step 3 is utter nonsense as written (all the even numbers for which Goldbach is true being less than whatever) since there are tons of even numbers bigger than 4*10^18 that are sums of two primes, like say $p+3$, $p$ arbitrary prime going to infinity; this is not even the type of subtle fallacy that appears in the RH case and requires thought to figure out, but complete misunderstanding of elementary math.
$endgroup$
– Conrad
Mar 15 at 1:51