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Is the problem Find all $x,y in mathbfN$ such that $binomx2 = binomy5$ solved?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Solving quadratic Diophantine equations: $5n^2+2n+1=y^2$How to solve Linear Diophantine equations?How to think about minors of the rectangular matrix in the context of a system of Diophantine linear equationsNonlinear system Diophantus.How to solve these system of linear equations?How can this problem be solved using continued fractions?A Diophantine equation involving factorialSolution of Certain Number Theoretic Problems.On diophantine equations involving Stirling numbers of the second kind I: the equation $mbrace k-nbrace k=z^k$ for fixed integers $kgeq 3$










4












$begingroup$


I was recently browsing and came upon this document which gives some open problems involving Diophantine Equations.
Document: http://www.math.leidenuniv.nl/~evertse/07-workshop-problems.pdf



Upon searching a bit, I found that the first problem given has been solved. Is the second problem also solved?
The problem is Find all $x,y in mathbfN$ such that $binomx2 = binomy5$.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    It looks a bit misleading. The problem is to find all $x, y in mathbbN$ with $binomx2 = binomy5$. It sounds like you are claiming that they are equal for all $x, y$, which they're not of course.
    $endgroup$
    – Randall
    Apr 2 at 1:04










  • $begingroup$
    google.com/amp/www.algebra.com/algebra/homework/Permutations/…
    $endgroup$
    – lab bhattacharjee
    Apr 2 at 1:30










  • $begingroup$
    @Randall thanks for that - have edited the question accordingly....@lab bhattacharjee I fail to see what that link has to do with my question or am I misunderstanding something?
    $endgroup$
    – blue apples
    Apr 2 at 1:35







  • 2




    $begingroup$
    de Weger, Equal binomial coefficients: some elementary considerations, J. Number Thy. 63 (1997) 373-386, gives the nontrivial examples 78-choose-2 equals 15-choose-5 equals 3003, and 153-choose-2 equals 19-choose-5 equals 11628, and conjectures there are no others.
    $endgroup$
    – Gerry Myerson
    Apr 2 at 3:14















4












$begingroup$


I was recently browsing and came upon this document which gives some open problems involving Diophantine Equations.
Document: http://www.math.leidenuniv.nl/~evertse/07-workshop-problems.pdf



Upon searching a bit, I found that the first problem given has been solved. Is the second problem also solved?
The problem is Find all $x,y in mathbfN$ such that $binomx2 = binomy5$.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    It looks a bit misleading. The problem is to find all $x, y in mathbbN$ with $binomx2 = binomy5$. It sounds like you are claiming that they are equal for all $x, y$, which they're not of course.
    $endgroup$
    – Randall
    Apr 2 at 1:04










  • $begingroup$
    google.com/amp/www.algebra.com/algebra/homework/Permutations/…
    $endgroup$
    – lab bhattacharjee
    Apr 2 at 1:30










  • $begingroup$
    @Randall thanks for that - have edited the question accordingly....@lab bhattacharjee I fail to see what that link has to do with my question or am I misunderstanding something?
    $endgroup$
    – blue apples
    Apr 2 at 1:35







  • 2




    $begingroup$
    de Weger, Equal binomial coefficients: some elementary considerations, J. Number Thy. 63 (1997) 373-386, gives the nontrivial examples 78-choose-2 equals 15-choose-5 equals 3003, and 153-choose-2 equals 19-choose-5 equals 11628, and conjectures there are no others.
    $endgroup$
    – Gerry Myerson
    Apr 2 at 3:14













4












4








4





$begingroup$


I was recently browsing and came upon this document which gives some open problems involving Diophantine Equations.
Document: http://www.math.leidenuniv.nl/~evertse/07-workshop-problems.pdf



Upon searching a bit, I found that the first problem given has been solved. Is the second problem also solved?
The problem is Find all $x,y in mathbfN$ such that $binomx2 = binomy5$.










share|cite|improve this question











$endgroup$




I was recently browsing and came upon this document which gives some open problems involving Diophantine Equations.
Document: http://www.math.leidenuniv.nl/~evertse/07-workshop-problems.pdf



Upon searching a bit, I found that the first problem given has been solved. Is the second problem also solved?
The problem is Find all $x,y in mathbfN$ such that $binomx2 = binomy5$.







number-theory diophantine-equations combinatorial-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 2 at 1:34







blue apples

















asked Apr 2 at 1:02









blue applesblue apples

454




454







  • 1




    $begingroup$
    It looks a bit misleading. The problem is to find all $x, y in mathbbN$ with $binomx2 = binomy5$. It sounds like you are claiming that they are equal for all $x, y$, which they're not of course.
    $endgroup$
    – Randall
    Apr 2 at 1:04










  • $begingroup$
    google.com/amp/www.algebra.com/algebra/homework/Permutations/…
    $endgroup$
    – lab bhattacharjee
    Apr 2 at 1:30










  • $begingroup$
    @Randall thanks for that - have edited the question accordingly....@lab bhattacharjee I fail to see what that link has to do with my question or am I misunderstanding something?
    $endgroup$
    – blue apples
    Apr 2 at 1:35







  • 2




    $begingroup$
    de Weger, Equal binomial coefficients: some elementary considerations, J. Number Thy. 63 (1997) 373-386, gives the nontrivial examples 78-choose-2 equals 15-choose-5 equals 3003, and 153-choose-2 equals 19-choose-5 equals 11628, and conjectures there are no others.
    $endgroup$
    – Gerry Myerson
    Apr 2 at 3:14












  • 1




    $begingroup$
    It looks a bit misleading. The problem is to find all $x, y in mathbbN$ with $binomx2 = binomy5$. It sounds like you are claiming that they are equal for all $x, y$, which they're not of course.
    $endgroup$
    – Randall
    Apr 2 at 1:04










  • $begingroup$
    google.com/amp/www.algebra.com/algebra/homework/Permutations/…
    $endgroup$
    – lab bhattacharjee
    Apr 2 at 1:30










  • $begingroup$
    @Randall thanks for that - have edited the question accordingly....@lab bhattacharjee I fail to see what that link has to do with my question or am I misunderstanding something?
    $endgroup$
    – blue apples
    Apr 2 at 1:35







  • 2




    $begingroup$
    de Weger, Equal binomial coefficients: some elementary considerations, J. Number Thy. 63 (1997) 373-386, gives the nontrivial examples 78-choose-2 equals 15-choose-5 equals 3003, and 153-choose-2 equals 19-choose-5 equals 11628, and conjectures there are no others.
    $endgroup$
    – Gerry Myerson
    Apr 2 at 3:14







1




1




$begingroup$
It looks a bit misleading. The problem is to find all $x, y in mathbbN$ with $binomx2 = binomy5$. It sounds like you are claiming that they are equal for all $x, y$, which they're not of course.
$endgroup$
– Randall
Apr 2 at 1:04




$begingroup$
It looks a bit misleading. The problem is to find all $x, y in mathbbN$ with $binomx2 = binomy5$. It sounds like you are claiming that they are equal for all $x, y$, which they're not of course.
$endgroup$
– Randall
Apr 2 at 1:04












$begingroup$
google.com/amp/www.algebra.com/algebra/homework/Permutations/…
$endgroup$
– lab bhattacharjee
Apr 2 at 1:30




$begingroup$
google.com/amp/www.algebra.com/algebra/homework/Permutations/…
$endgroup$
– lab bhattacharjee
Apr 2 at 1:30












$begingroup$
@Randall thanks for that - have edited the question accordingly....@lab bhattacharjee I fail to see what that link has to do with my question or am I misunderstanding something?
$endgroup$
– blue apples
Apr 2 at 1:35





$begingroup$
@Randall thanks for that - have edited the question accordingly....@lab bhattacharjee I fail to see what that link has to do with my question or am I misunderstanding something?
$endgroup$
– blue apples
Apr 2 at 1:35





2




2




$begingroup$
de Weger, Equal binomial coefficients: some elementary considerations, J. Number Thy. 63 (1997) 373-386, gives the nontrivial examples 78-choose-2 equals 15-choose-5 equals 3003, and 153-choose-2 equals 19-choose-5 equals 11628, and conjectures there are no others.
$endgroup$
– Gerry Myerson
Apr 2 at 3:14




$begingroup$
de Weger, Equal binomial coefficients: some elementary considerations, J. Number Thy. 63 (1997) 373-386, gives the nontrivial examples 78-choose-2 equals 15-choose-5 equals 3003, and 153-choose-2 equals 19-choose-5 equals 11628, and conjectures there are no others.
$endgroup$
– Gerry Myerson
Apr 2 at 3:14










1 Answer
1






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oldest

votes


















4












$begingroup$

According to Blokhuis, Brouwer, and de Weger, Binomial collisions and near collisions, Integers 17 (2017) #A64, the question was settled in Bugeaud, Mignotte, Siksek, Stoll, and Tengely, Integral points on hyperelliptic curves, Algebra Number Theory 2 (2008) 859-885; there are no nontrivial solutions, other than those given in the comment.






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    $begingroup$

    According to Blokhuis, Brouwer, and de Weger, Binomial collisions and near collisions, Integers 17 (2017) #A64, the question was settled in Bugeaud, Mignotte, Siksek, Stoll, and Tengely, Integral points on hyperelliptic curves, Algebra Number Theory 2 (2008) 859-885; there are no nontrivial solutions, other than those given in the comment.






    share|cite|improve this answer









    $endgroup$

















      4












      $begingroup$

      According to Blokhuis, Brouwer, and de Weger, Binomial collisions and near collisions, Integers 17 (2017) #A64, the question was settled in Bugeaud, Mignotte, Siksek, Stoll, and Tengely, Integral points on hyperelliptic curves, Algebra Number Theory 2 (2008) 859-885; there are no nontrivial solutions, other than those given in the comment.






      share|cite|improve this answer









      $endgroup$















        4












        4








        4





        $begingroup$

        According to Blokhuis, Brouwer, and de Weger, Binomial collisions and near collisions, Integers 17 (2017) #A64, the question was settled in Bugeaud, Mignotte, Siksek, Stoll, and Tengely, Integral points on hyperelliptic curves, Algebra Number Theory 2 (2008) 859-885; there are no nontrivial solutions, other than those given in the comment.






        share|cite|improve this answer









        $endgroup$



        According to Blokhuis, Brouwer, and de Weger, Binomial collisions and near collisions, Integers 17 (2017) #A64, the question was settled in Bugeaud, Mignotte, Siksek, Stoll, and Tengely, Integral points on hyperelliptic curves, Algebra Number Theory 2 (2008) 859-885; there are no nontrivial solutions, other than those given in the comment.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 2 at 3:27









        Gerry MyersonGerry Myerson

        148k8152306




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