Let, $xy=n^{\underline k} = n(n-1)(n-2)\cdot\ldots\cdot (n-k+1)$ and it is given that $ \gcd(x,y)=1$ with one of $x,y$ is odd, another is even. When does $\gcd (x-1,y-1)=z>1$?

In other words, what are the necessary (non-trivial) or necessary and sufficient condition(s) on variables $n,k,x,y$ for such cases?

Examples of trivial cases:

1. $z |(x − 1) +(y − 1) = x+y-2 \implies x+y \equiv2 \pmod z$.
2. A sufficient condition would be that $x−1∣y−x$.

A similar-looking problem:

Let $m, n$ be distinct positive integers. Prove that $\gcd(m, n) + \gcd(m + 1, n + 1) + \gcd(m + 2, n + 2) ≤ 2|m − n| + 1$. Further, determine when equality holds (from 34th Indian National Mathematical Olympiad-2019  ).

I mentioned above problem just to show this kind of problem exists, and the problem I posted is not a random problem.

A related research result:

The below result is found in the article entitled "On the factorization of consecutive integers" by M.A. Bennett, M. Filaseta and O. Trifonov on February 26, 2007 -

Theorem 1.1. If $k ∈ {5, 7}, n ≥ 2k$ is an integer, and we write $\binom nk = U · V$, where $U$ and $V$ are integers with $P (U) ≤ k$ and $V$ coprime to $k!$, then it follows that $V > U$, unless $(n, k) ∈ {(10, 5), (12, 5), (21, 7), (28, 5), (30, 7), (54, 7)} $.

Note:

Please consider non-trivial cases.

Let, $xy=n^{\underline k} = n(n-1)(n-2)\cdot\dotsm\cdot (n-k+1)$ and it is given that $ \gcd(x,y)=1$ with one of $x$, $y$ is odd, another is even. When is $\gcd (x-1,y-1)=z>1$?

In other words, what are necessary (non-trivial) or necessary and sufficient condition(s) on variables $n$, $k$, $x$, $y$ for such cases?

Examples of trivial cases:

1. $z \mid (x − 1) +(y − 1) = x+y-2 \implies x+y \equiv2 \pmod z$.
2. A sufficient condition would be that $x−1\mid y−x$.

A similar-looking problem:

Let $m$, $n$ be distinct positive integers. Prove that $\gcd(m, n) + \gcd(m + 1, n + 1) + \gcd(m + 2, n + 2) \le 2\lvert m − n\rvert + 1$. Further, determine when equality holds (from 34th Indian National Mathematical Olympiad-2019).

I mentioned the above problem just to show this kind of problem exists, and the problem I posted is not a random problem.

A related research result:

The below result is found in the article entitled "On the factorization of consecutive integers" by M. A. Bennett, M. Filaseta and O. Trifonov on February 26, 2007.

Theorem 1.1. If $k \in \{5, 7\}$, $n \ge 2k$ is an integer, and we write $\binom nk = U \cdot V$, where $U$ and $V$ are integers where the greatest prime factor $P (U)$ of $U$ is $\le k$ and $V$ is coprime to $k!$, then it follows that $V > U$, unless $(n, k) \in \{(10, 5), (12, 5), (21, 7), (28, 5), (30, 7), (54, 7)\} $.

Note:

Please consider non-trivial cases.

Let, $xy=n^{\underline k} = n(n-1)(n-2)\cdot\ldots\cdot (n-k+1)$ and it is given that $ \gcd(x,y)=1$ with one of $x,y$ is odd, another is even. When does $\gcd (x-1,y-1)=z>1$?

In other words, what are the necessary (non-trivial) or necessary and sufficient condition(s) on variables $n,k,x,y$ for such cases?

Examples of Trivial Cases:

1. $z |(x − 1) +(y − 1) = x+y-2 \implies x+y \equiv2 \pmod z$.
2. A sufficient condition would be that $x−1∣y−x$.

A Similar-Looking Problem:

Let $m, n$ be distinct positive integers. Prove that $\gcd(m, n) + \gcd(m + 1, n + 1) + \gcd(m + 2, n + 2) ≤ 2|m − n| + 1$. Further, determine when equality holds (from 34th Indian National Mathematical Olympiad-2019 ).

I mentioned above problem just to show this kind of problem exists, and the problem I posted is not a random problem.

A Related Research Result:

The below result is found in the article entitled "On the factorization of consecutive integers" by M.A. Bennett, M. Filaseta and O. Trifonov on February 26, 2007 -

Theorem 1.1. If $k ∈ {5, 7}, n ≥ 2k$ is an integer, and we write $\binom nk = U · V$, where $U$ and $V$ are integers with $P (U) ≤ k$ and $V$ coprime to $k!$, then it follows that $V > U$, unless $(n, k) ∈ {(10, 5), (12, 5), (21, 7), (28, 5), (30, 7), (54, 7)} $.

Note:

Please consider non-trivial cases.

Let, $xy=n^{\underline k} = n(n-1)(n-2)\cdot\ldots\cdot (n-k+1)$ and it is given that $ \gcd(x,y)=1$ with one of $x,y$ is odd, another is even. When does $\gcd (x-1,y-1)=z>1$?

In other words, what are the necessary (non-trivial) or necessary and sufficient condition(s) on variables $n,k,x,y$ for such cases?

Examples of trivial cases:

1. $z |(x − 1) +(y − 1) = x+y-2 \implies x+y \equiv2 \pmod z$.
2. A sufficient condition would be that $x−1∣y−x$.

A similar-looking problem:

Let $m, n$ be distinct positive integers. Prove that $\gcd(m, n) + \gcd(m + 1, n + 1) + \gcd(m + 2, n + 2) ≤ 2|m − n| + 1$. Further, determine when equality holds (from 34th Indian National Mathematical Olympiad-2019 ).

I mentioned above problem just to show this kind of problem exists, and the problem I posted is not a random problem.

A related research result:

The below result is found in the article entitled "On the factorization of consecutive integers" by M.A. Bennett, M. Filaseta and O. Trifonov on February 26, 2007 -

Theorem 1.1. If $k ∈ {5, 7}, n ≥ 2k$ is an integer, and we write $\binom nk = U · V$, where $U$ and $V$ are integers with $P (U) ≤ k$ and $V$ coprime to $k!$, then it follows that $V > U$, unless $(n, k) ∈ {(10, 5), (12, 5), (21, 7), (28, 5), (30, 7), (54, 7)} $.

Note:

Please consider non-trivial cases.

Let, $xy=n^{\underline k} = n(n-1)(n-2)\cdot\ldots\cdot (n-+1)$ and it is given that $ \gcd(x,y)=1$ with one of $x,y$ is odd, another is even. When does $\gcd (x-1,y-1)=z>1$?

In other words, what are the necessary (non-trivial) or necessary and sufficient condition(s) on variables $n,k,x,y$ for such cases?

Examples of Trivial Cases:

1. $z |(x − 1) +(y − 1) = x+y-2 \implies x+y \equiv2 \pmod z$.
2. A sufficient condition would be that $x−1∣y−x$.

A Similar-Looking Problem:

Let $m, n$ be distinct positive integers. Prove that $\gcd(m, n) + \gcd(m + 1, n + 1) + \gcd(m + 2, n + 2) ≤ 2|m − n| + 1$. Further, determine when equality holds (from 34th Indian National Mathematical Olympiad-2019 ).

I mentioned above problem just to show this kind of problem exists, and the problem I posted is not a random problem.

A Related Research Result:

The below result is found in the article entitled "On the factorization of consecutive integers" by M.A. Bennett, M. Filaseta and O. Trifonov on February 26, 2007 -

Theorem 1.1. If $k ∈ {5, 7}, n ≥ 2k$ is an integer, and we write $\binom nk = U · V$, where $U$ and $V$ are integers with $P (U) ≤ k$ and $V$ coprime to $k!$, then it follows that $V > U$, unless $(n, k) ∈ {(10, 5), (12, 5), (21, 7), (28, 5), (30, 7), (54, 7)} $.

Note:

Please consider non-trivial cases.

Let, $xy=n^{\underline k} = n(n-1)(n-2)\cdot\ldots\cdot (n-k+1)$ and it is given that $ \gcd(x,y)=1$ with one of $x,y$ is odd, another is even. When does $\gcd (x-1,y-1)=z>1$?

In other words, what are the necessary (non-trivial) or necessary and sufficient condition(s) on variables $n,k,x,y$ for such cases?

Examples of Trivial Cases:

1. $z |(x − 1) +(y − 1) = x+y-2 \implies x+y \equiv2 \pmod z$.
2. A sufficient condition would be that $x−1∣y−x$.

A Similar-Looking Problem:

Let $m, n$ be distinct positive integers. Prove that $\gcd(m, n) + \gcd(m + 1, n + 1) + \gcd(m + 2, n + 2) ≤ 2|m − n| + 1$. Further, determine when equality holds (from 34th Indian National Mathematical Olympiad-2019 ).

I mentioned above problem just to show this kind of problem exists, and the problem I posted is not a random problem.

A Related Research Result:

The below result is found in the article entitled "On the factorization of consecutive integers" by M.A. Bennett, M. Filaseta and O. Trifonov on February 26, 2007 -

Theorem 1.1. If $k ∈ {5, 7}, n ≥ 2k$ is an integer, and we write $\binom nk = U · V$, where $U$ and $V$ are integers with $P (U) ≤ k$ and $V$ coprime to $k!$, then it follows that $V > U$, unless $(n, k) ∈ {(10, 5), (12, 5), (21, 7), (28, 5), (30, 7), (54, 7)} $.

Note:

Please consider non-trivial cases.