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Harpo Marx provided a great answer to this question based on the Thue equation.

I investigated the Thue equation and while User 631's argument was not complete, it does appear possible to make it complete. I am very glad to be corrected if I have make a mistake in my reasoning.

Here's the argument as I understand it:

(1) For any prime $p$, there is a finite number of combinations of primes that are less than $p$.

(2) For any of these combinations there exists an integer $x$ such that if a combination is greater than $x$, then at least one of the primes that make up the combination are of a degree greater than $2$.

(3) Let $c$ be either the highest of the values $x$ from step (2) or $4p^{4}(3\prod_{p} p^{\frac{1}{2}})^3$ depending on which is higher.

(4) There is a finite number of ways that we can pair these different combinations so that we have an equation of the form $ax^3 - by^3 = c$ where $c < p$.

(5) If $a \ne b$ and $gcd(x,y)=1$, then using a result from Siegel, there is at most $1$ solution (see Theorem A in the link below).

http://matwbn.icm.edu.pl/ksiazki/aa/aa75/aa7538.pdf

(6) if $a = b$ and $a = c$, then using a result from Michael Bennett, there is at most $1$ solution. Here's the reference:

M. A. Bennett, Rational approximation to algebraic numbers of small height : the Diophantine equation $|ax^n-by^n|=1$}, J. Reine Angew. Math. 535 (2001), 1-49

http://www.math.ubc.ca/~bennett/B-Crelle2.pdf

(7) if $a = b$ and $a < c$, then the equation has a form such as:

$x^3 - y^3 = \frac{c}{a}$

This is a Thue Equation and we can conclude that there is a finite number of solutions.

(8) if $a \ne b$ and $gcd(x,y) > 1$, then it means that both combinations consist of the same prime so we have an equation of the form:

$x^m - x^n = c$

Then, as I understand it, we have a Thue equation so we can again assume that there is a finite number solutions.

I believe that covers all the possible cases.

Since, there are only a finite number of solutions, it follows that there exists an integer $c$ which is greater than all of these solutions and for all $x \ge c$, we have at most $1$ integer in the sequence where $gpc(x+i) \le p$.

Apologies for the length of this argument. I'm sure a professional mathematician such as user 631 would be able to state the argument more elegantly. :-)

I heard a great answer to this question based on the Thue equation.

I investigated the Thue equation and there was one point that was not clear to me. It seems to me that there are an infinite number of values that $a$ and $b$ can take. If there are an infinite number of combinations of finite solutions, then there is an infinite number of solutions. Right?

So, if I understand it, the Thue equation alone doesn't seem to work. I apologize if I am misunderstanding the classical result there.

Here's an argument that seems to work as far as I understand:

(1) For any prime $p$, there is a finite number of combinations of primes that are less than $p$.

(2) For any of these combinations there exists an integer $x$ such that if a combination is greater than $x$, then at least one of the primes that make up the combination are of a degree greater than $2$.

(3) Let $c$ be either the highest of the values $x$ from step (2) or $4p^{4}(3\prod_{p} p^{\frac{1}{2}})^3$ depending on which is higher.

(4) There is a finite number of ways that we can pair these different combinations so that we have an equation of the form $ax^3 - by^3 = c$ where $c < p$.

(5) If $a \ne b$ and $gcd(x,y)=1$, then using a result from Siegel, there is at most $1$ solution (see Theorem A in the link below).

http://matwbn.icm.edu.pl/ksiazki/aa/aa75/aa7538.pdf

(6) if $a = b$ and $a = c$, then using a result from Michael Bennett, there is at most $1$ solution. Here's the reference:

M. A. Bennett, Rational approximation to algebraic numbers of small height : the Diophantine equation $|ax^n-by^n|=1$}, J. Reine Angew. Math. 535 (2001), 1-49

http://www.math.ubc.ca/~bennett/B-Crelle2.pdf

(7) if $a = b$ and $a < c$, then the equation has a form such as:

$x^3 - y^3 = \frac{c}{a}$

This is a Thue Equation and we can conclude that there is a finite number of solutions.

(8) if $a \ne b$ and $gcd(x,y) > 1$, then it means that both combinations consist of the same prime so we have an equation of the form:

$x^m - x^n = c$

Then, as I understand it, we have a Thue equation so we can again assume that there is a finite number solutions.

I believe that covers all the possible cases.

Since, there are only a finite number of solutions, it follows that there exists an integer $c$ which is greater than all of these solutions and for all $x \ge c$, we have at most $1$ integer in the sequence where $gpc(x+i) \le p$.

Apologies for the length of this argument. I'm sure a professional mathematician such as user 631 would be able to state the argument more elegantly. :-)

User 631 ( http://mathoverflow.net/users/631 ) provided an answer to this question based on the Thue equation and then he deleted his comment.

I investigated the Thue equation and while User 631's argument was not complete, it does appear possible to make it complete. I am very glad to be corrected if I have make a mistake in my reasoning.

The Thue equation alone doesn't resolve this question but with the addition of some other results, User 631's line of reasoning seems to me to work.

User 631, please post an additional comment if I am misinterpreting your reasons for deleting your observation about the Thue Equation.

Here's the argument as I understand it:

(1) For any prime $p$, there is a finite number of combinations of primes that are less than $p$.

(2) For any of these combinations there exists an integer $x$ such that if a combination is greater than $x$, then at least one of the primes that make up the combination are of a degree greater than $2$.

(3) Let $c$ be either the highest of the values $x$ from step (2) or $4p^{4}(3\prod_{p} p^{\frac{1}{2}})^3$ depending on which is higher.

(4) There is a finite number of ways that we can pair these different combinations so that we have an equation of the form $ax^3 - by^3 = c$ where $c < p$.

(5) If $a \ne b$ and $gcd(x,y)=1$, then using a result from Siegel, there is at most $1$ solution (see Theorem A in the link below).

http://matwbn.icm.edu.pl/ksiazki/aa/aa75/aa7538.pdf

(6) if $a = b$ and $a = c$, then using a result from Michael Bennett, there is at most $1$ solution. Here's the reference:

M. A. Bennett, Rational approximation to algebraic numbers of small height : the Diophantine equation $|ax^n-by^n|=1$}, J. Reine Angew. Math. 535 (2001), 1-49

http://www.math.ubc.ca/~bennett/B-Crelle2.pdf

(7) if $a = b$ and $a < c$, then the equation has a form such as:

$x^3 - y^3 = \frac{c}{a}$

This is a Thue Equation and we can conclude that there is a finite number of solutions.

(8) if $a \ne b$ and $gcd(x,y) > 1$, then it means that both combinations consist of the same prime so we have an equation of the form:

$x^m - x^n = c$

Then, as I understand it, we have a Thue equation so we can again assume that there is a finite number solutions.

I believe that covers all the possible cases.

Since, there are only a finite number of solutions, it follows that there exists an integer $c$ which is greater than all of these solutions and for all $x \ge c$, we have at most $1$ integer in the sequence where $gpc(x+i) \le p$.

Apologies for the length of this argument. I'm sure a professional mathematician such as user 631 would be able to state the argument more elegantly. :-)

Harpo Marx provided a great answer to this question based on the Thue equation.

I investigated the Thue equation and while User 631's argument was not complete, it does appear possible to make it complete. I am very glad to be corrected if I have make a mistake in my reasoning.

Here's the argument as I understand it:

(1) For any prime $p$, there is a finite number of combinations of primes that are less than $p$.

(2) For any of these combinations there exists an integer $x$ such that if a combination is greater than $x$, then at least one of the primes that make up the combination are of a degree greater than $2$.

(3) Let $c$ be either the highest of the values $x$ from step (2) or $4p^{4}(3\prod_{p} p^{\frac{1}{2}})^3$ depending on which is higher.

(4) There is a finite number of ways that we can pair these different combinations so that we have an equation of the form $ax^3 - by^3 = c$ where $c < p$.

(5) If $a \ne b$ and $gcd(x,y)=1$, then using a result from Siegel, there is at most $1$ solution (see Theorem A in the link below).

http://matwbn.icm.edu.pl/ksiazki/aa/aa75/aa7538.pdf

(6) if $a = b$ and $a = c$, then using a result from Michael Bennett, there is at most $1$ solution. Here's the reference:

M. A. Bennett, Rational approximation to algebraic numbers of small height : the Diophantine equation $|ax^n-by^n|=1$}, J. Reine Angew. Math. 535 (2001), 1-49

http://www.math.ubc.ca/~bennett/B-Crelle2.pdf

(7) if $a = b$ and $a < c$, then the equation has a form such as:

$x^3 - y^3 = \frac{c}{a}$

This is a Thue Equation and we can conclude that there is a finite number of solutions.

(8) if $a \ne b$ and $gcd(x,y) > 1$, then it means that both combinations consist of the same prime so we have an equation of the form:

$x^m - x^n = c$

Then, as I understand it, we have a Thue equation so we can again assume that there is a finite number solutions.

I believe that covers all the possible cases.

Since, there are only a finite number of solutions, it follows that there exists an integer $c$ which is greater than all of these solutions and for all $x \ge c$, we have at most $1$ integer in the sequence where $gpc(x+i) \le p$.

Apologies for the length of this argument. I'm sure a professional mathematician such as user 631 would be able to state the argument more elegantly. :-)

User 631 ( http://mathoverflow.net/users/631 ) provided an answer to this question based on the Thue equation and then he deleted his comment.

I investigated the Thue equation and while User 631's argument was not complete, it does appear possible to make it complete. I am very glad to be corrected if I have made a mistake in my reasoning.

The Thue equation alone doesn't resolve this question but with the addition of some other results, User 631's line of reasoning seems to me to work.

User 631, please post an additional comment if I am misinterpreting your reasons for deleting your observation about the Thue Equation.

Here's the argument as I understand it:

(1) For any prime $p$, there is a finite number of combinations of primes that are less than $p$.

(2) For any of these combinations there exists an integer $x$ such that if a combination is greater than $x$, then at least of the primes that make up the combination are of a degree greater than $2$.

(3) Let $c$ be either the highest of the values $x$ from step (2) or $4p^{4}(3\prod_{p} p^{\frac{1}{2}})^3$ depending on which is higher.

(4) There is a finite number of ways that we can pair these different combinations so that we have an equation of the form $ax^3 - by^3 = c$ where $c < p$.

(5) If $a \ne b$ and $gcd(x,y)=1$, then using a result from Siegel, there is at most $1$ solution (see Theorem A in the link below).

http://matwbn.icm.edu.pl/ksiazki/aa/aa75/aa7538.pdf

(6) if $a = b$ and $a = c$, then using a result from Michael Bennett, there is at most $1$ solution. Here's the reference:

M. A. Bennett, Rational approximation to algebraic numbers of small height : the Diophantine equation $|ax^n-by^n|=1$}, J. Reine Angew. Math. 535 (2001), 1-49

http://www.math.ubc.ca/~bennett/B-Crelle2.pdf

(7) if $a = b$ and $a < c$, then the equation has a form such as:

$x^3 - y^3 = \frac{c}{a}$

This is a Thue Equation and we can conclude that there is a finite number of solutions.

(8) if $a \ne b$ and $gcd(x,y) > 1$, then it means that both combinations consist of the same prime so we have an equation of the form:

$x^m - x^n = c$

Then, as I understand it, we have a Thue equation so we can again assume that there is a finite number solutions.

I believe that covers all the possible cases.

Since, there are only a finite number of solutions, it follows that there exists an integer $c$ which is greater than all of these solutions and for all $x \ge c$, we have at most $1$ integer in the sequence where $gpc(x+i) \le p$.

Apologies for the length of this argument. I'm sure an professional mathematician such as user 631 would be able to state the argument more elegantly. :-)

User 631 ( http://mathoverflow.net/users/631 ) provided an answer to this question based on the Thue equation and then he deleted his comment.

I investigated the Thue equation and while User 631's argument was not complete, it does appear possible to make it complete. I am very glad to be corrected if I have make a mistake in my reasoning.

The Thue equation alone doesn't resolve this question but with the addition of some other results, User 631's line of reasoning seems to me to work.

User 631, please post an additional comment if I am misinterpreting your reasons for deleting your observation about the Thue Equation.

Here's the argument as I understand it:

(1) For any prime $p$, there is a finite number of combinations of primes that are less than $p$.

(2) For any of these combinations there exists an integer $x$ such that if a combination is greater than $x$, then at least one of the primes that make up the combination are of a degree greater than $2$.

(3) Let $c$ be either the highest of the values $x$ from step (2) or $4p^{4}(3\prod_{p} p^{\frac{1}{2}})^3$ depending on which is higher.

(4) There is a finite number of ways that we can pair these different combinations so that we have an equation of the form $ax^3 - by^3 = c$ where $c < p$.

(5) If $a \ne b$ and $gcd(x,y)=1$, then using a result from Siegel, there is at most $1$ solution (see Theorem A in the link below).

http://matwbn.icm.edu.pl/ksiazki/aa/aa75/aa7538.pdf

(6) if $a = b$ and $a = c$, then using a result from Michael Bennett, there is at most $1$ solution. Here's the reference:

M. A. Bennett, Rational approximation to algebraic numbers of small height : the Diophantine equation $|ax^n-by^n|=1$}, J. Reine Angew. Math. 535 (2001), 1-49

http://www.math.ubc.ca/~bennett/B-Crelle2.pdf

(7) if $a = b$ and $a < c$, then the equation has a form such as:

$x^3 - y^3 = \frac{c}{a}$

This is a Thue Equation and we can conclude that there is a finite number of solutions.

(8) if $a \ne b$ and $gcd(x,y) > 1$, then it means that both combinations consist of the same prime so we have an equation of the form:

$x^m - x^n = c$

Then, as I understand it, we have a Thue equation so we can again assume that there is a finite number solutions.

I believe that covers all the possible cases.

Since, there are only a finite number of solutions, it follows that there exists an integer $c$ which is greater than all of these solutions and for all $x \ge c$, we have at most $1$ integer in the sequence where $gpc(x+i) \le p$.

Apologies for the length of this argument. I'm sure a professional mathematician such as user 631 would be able to state the argument more elegantly. :-)