Sizes of a pair of sequences with identical sums of pairs












4












$begingroup$


I am currently collecting various problems for an exam for my students. While looking through old homework assignments of my colleagues I came upon the following problem (marked as difficult):



Given two sequences of natural numbers ${a_k}$ and ${b_k}$, $k=1,ldots,n$ (with non-identical sets of elements) such that the sets of their pairwise sums $${a_1+a_2,a_1 + a_3,ldots, a_{n-1}+a_n}$$ and $${b_1+b_2,b_1 + b_3,ldots, b_{n-1}+b_n}$$ coincide, show that $n=2^m, minmathbb{N}.$



Of course, I am not going to assign a problem I couldn't solve myself to the students, but I would like to see a solution to this. This problem was accompanied with the following tip:



"Use the fact that if for two polynomials $F(x)$ and $G(x)$ if $F(1)=G(1)$, then $F(x)-G(x)=(x-1)^kH(x)$, where $H(1)neq 0$".






combinatorics discrete-mathematics







share|cite|improve this question









$endgroup$












  • $begingroup$
    What course was the homework for?
    $endgroup$
    – Peter Taylor
    Dec 8 '18 at 21:14










  • $begingroup$
    @PeterTaylor It is called "Boolean algebra, combinatorics and graph theory", it is an introductory course for first year students.
    $endgroup$
    – Serg
    Dec 8 '18 at 21:32










  • $begingroup$
    Can you give an example of such $a_k,b_k$ with say $n=4$? I realize it's just to show it can't happen for $n$ not a power of 2, but was wondering what kind of examples might be for small powers of $2.$
    $endgroup$
    – coffeemath
    Dec 8 '18 at 21:47
















4












$begingroup$


I am currently collecting various problems for an exam for my students. While looking through old homework assignments of my colleagues I came upon the following problem (marked as difficult):



Given two sequences of natural numbers ${a_k}$ and ${b_k}$, $k=1,ldots,n$ (with non-identical sets of elements) such that the sets of their pairwise sums $${a_1+a_2,a_1 + a_3,ldots, a_{n-1}+a_n}$$ and $${b_1+b_2,b_1 + b_3,ldots, b_{n-1}+b_n}$$ coincide, show that $n=2^m, minmathbb{N}.$



Of course, I am not going to assign a problem I couldn't solve myself to the students, but I would like to see a solution to this. This problem was accompanied with the following tip:



"Use the fact that if for two polynomials $F(x)$ and $G(x)$ if $F(1)=G(1)$, then $F(x)-G(x)=(x-1)^kH(x)$, where $H(1)neq 0$".






combinatorics discrete-mathematics







share|cite|improve this question









$endgroup$












  • $begingroup$
    What course was the homework for?
    $endgroup$
    – Peter Taylor
    Dec 8 '18 at 21:14










  • $begingroup$
    @PeterTaylor It is called "Boolean algebra, combinatorics and graph theory", it is an introductory course for first year students.
    $endgroup$
    – Serg
    Dec 8 '18 at 21:32










  • $begingroup$
    Can you give an example of such $a_k,b_k$ with say $n=4$? I realize it's just to show it can't happen for $n$ not a power of 2, but was wondering what kind of examples might be for small powers of $2.$
    $endgroup$
    – coffeemath
    Dec 8 '18 at 21:47














4












4








4


2



$begingroup$


I am currently collecting various problems for an exam for my students. While looking through old homework assignments of my colleagues I came upon the following problem (marked as difficult):



Given two sequences of natural numbers ${a_k}$ and ${b_k}$, $k=1,ldots,n$ (with non-identical sets of elements) such that the sets of their pairwise sums $${a_1+a_2,a_1 + a_3,ldots, a_{n-1}+a_n}$$ and $${b_1+b_2,b_1 + b_3,ldots, b_{n-1}+b_n}$$ coincide, show that $n=2^m, minmathbb{N}.$



Of course, I am not going to assign a problem I couldn't solve myself to the students, but I would like to see a solution to this. This problem was accompanied with the following tip:



"Use the fact that if for two polynomials $F(x)$ and $G(x)$ if $F(1)=G(1)$, then $F(x)-G(x)=(x-1)^kH(x)$, where $H(1)neq 0$".






combinatorics discrete-mathematics







share|cite|improve this question









$endgroup$




I am currently collecting various problems for an exam for my students. While looking through old homework assignments of my colleagues I came upon the following problem (marked as difficult):



Given two sequences of natural numbers ${a_k}$ and ${b_k}$, $k=1,ldots,n$ (with non-identical sets of elements) such that the sets of their pairwise sums $${a_1+a_2,a_1 + a_3,ldots, a_{n-1}+a_n}$$ and $${b_1+b_2,b_1 + b_3,ldots, b_{n-1}+b_n}$$ coincide, show that $n=2^m, minmathbb{N}.$



Of course, I am not going to assign a problem I couldn't solve myself to the students, but I would like to see a solution to this. This problem was accompanied with the following tip:



"Use the fact that if for two polynomials $F(x)$ and $G(x)$ if $F(1)=G(1)$, then $F(x)-G(x)=(x-1)^kH(x)$, where $H(1)neq 0$".






combinatorics discrete-mathematics




combinatorics discrete-mathematics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 20:37









SergSerg

611415




611415












  • $begingroup$
    What course was the homework for?
    $endgroup$
    – Peter Taylor
    Dec 8 '18 at 21:14










  • $begingroup$
    @PeterTaylor It is called "Boolean algebra, combinatorics and graph theory", it is an introductory course for first year students.
    $endgroup$
    – Serg
    Dec 8 '18 at 21:32










  • $begingroup$
    Can you give an example of such $a_k,b_k$ with say $n=4$? I realize it's just to show it can't happen for $n$ not a power of 2, but was wondering what kind of examples might be for small powers of $2.$
    $endgroup$
    – coffeemath
    Dec 8 '18 at 21:47


















  • $begingroup$
    What course was the homework for?
    $endgroup$
    – Peter Taylor
    Dec 8 '18 at 21:14










  • $begingroup$
    @PeterTaylor It is called "Boolean algebra, combinatorics and graph theory", it is an introductory course for first year students.
    $endgroup$
    – Serg
    Dec 8 '18 at 21:32










  • $begingroup$
    Can you give an example of such $a_k,b_k$ with say $n=4$? I realize it's just to show it can't happen for $n$ not a power of 2, but was wondering what kind of examples might be for small powers of $2.$
    $endgroup$
    – coffeemath
    Dec 8 '18 at 21:47
















$begingroup$
What course was the homework for?
$endgroup$
– Peter Taylor
Dec 8 '18 at 21:14




$begingroup$
What course was the homework for?
$endgroup$
– Peter Taylor
Dec 8 '18 at 21:14












$begingroup$
@PeterTaylor It is called "Boolean algebra, combinatorics and graph theory", it is an introductory course for first year students.
$endgroup$
– Serg
Dec 8 '18 at 21:32




$begingroup$
@PeterTaylor It is called "Boolean algebra, combinatorics and graph theory", it is an introductory course for first year students.
$endgroup$
– Serg
Dec 8 '18 at 21:32












$begingroup$
Can you give an example of such $a_k,b_k$ with say $n=4$? I realize it's just to show it can't happen for $n$ not a power of 2, but was wondering what kind of examples might be for small powers of $2.$
$endgroup$
– coffeemath
Dec 8 '18 at 21:47




$begingroup$
Can you give an example of such $a_k,b_k$ with say $n=4$? I realize it's just to show it can't happen for $n$ not a power of 2, but was wondering what kind of examples might be for small powers of $2.$
$endgroup$
– coffeemath
Dec 8 '18 at 21:47










2 Answers
2






active

oldest

votes


















5












$begingroup$

I will show that the statement is true with the condition that




  • The two multisets $A={a_i+a_j:1 le i<jle n}$ and
    $B={b_i+b_j:1 le i<jle n}$ are the same
    (i.e. if $x in A$ appears $k$ times in $A$ then
    $x$ appears $k$ times in $B$).


  • Two sets ${a_i}$ and ${b_i}$ are not the same.



Let $A(x)=sum_{i=1}^n x^{a_i}$ and $B(x)=sum_{i=1}^n x^{b_i}$
then we have $A^2(x)=A(x^2)+2sum_{iin A} x^i$ and similarly,
$B^2(x)=B(x^2)+2sum_{iin B} x^i$. Therefore,
$$A(x^2)-B(x^2)=A^2(x)-B^2(x)=[A(x)+B(x)][A(x)-B(x)].$$
Since $A(1)=B(1)=n$ so $A(x)-B(x)=(x-1)^kG(x)$ where $G(1)ne 0,
k ge 1$.This follows
$$(x^2-1)^k G(x^2)=A(x^2)-B(x^2)=[A(x)+B(x)]cdot (x-1)^k G(x).$$
Therefore, $(x+1)^k G(x^2)=[A(x)+B(x)]G(x)$. Substituting $x=1$ into this
and note that $A(1)+B(1)=2n$ and $G(1)ne 0$,
we obtain $n=2^{k-1}$, as desired.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Consider the sequence $(a_k)=(2,4,4)$ and the sequence $(b_k)=(3,3,5)$. Then the two sequences have non-identical sets of elements, but the sets of their pairwise sums are as follows:
    $$
    A={2+4,2+4,4+4}={6,8}
    $$
    $$
    B={3+3,3+5,3+5}={6,8}
    $$



    It would then seem that we need to alter the language to make the statement true... maybe it should be a multiset? Maybe the sequences can't have repeated values?



    EDIT: Some quick and dirty coding indicates the following question rewrite has a better chance of being true:



    Given two sets of natural numbers $A$ and $B$ where $Aneq B$ and $|A|=|B|=n$ such that the restricted sumsets $2^wedge A$ and $2^wedge B$ are equal, show that $n=2^m$ for some natural number $m$.



    To answer @coffeemath's comment in the OP, you can take $A={1,4,5,6}$ and $B={2,3,4,7}$ as examples of the case $n=4$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      RandomMathGuy--- Thanks for the $n=4$example!
      $endgroup$
      – coffeemath
      Dec 10 '18 at 0:14











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    I will show that the statement is true with the condition that




    • The two multisets $A={a_i+a_j:1 le i<jle n}$ and
      $B={b_i+b_j:1 le i<jle n}$ are the same
      (i.e. if $x in A$ appears $k$ times in $A$ then
      $x$ appears $k$ times in $B$).


    • Two sets ${a_i}$ and ${b_i}$ are not the same.



    Let $A(x)=sum_{i=1}^n x^{a_i}$ and $B(x)=sum_{i=1}^n x^{b_i}$
    then we have $A^2(x)=A(x^2)+2sum_{iin A} x^i$ and similarly,
    $B^2(x)=B(x^2)+2sum_{iin B} x^i$. Therefore,
    $$A(x^2)-B(x^2)=A^2(x)-B^2(x)=[A(x)+B(x)][A(x)-B(x)].$$
    Since $A(1)=B(1)=n$ so $A(x)-B(x)=(x-1)^kG(x)$ where $G(1)ne 0,
    k ge 1$    .This follows
    $$(x^2-1)^k G(x^2)=A(x^2)-B(x^2)=[A(x)+B(x)]cdot (x-1)^k G(x).$$
    Therefore, $(x+1)^k G(x^2)=[A(x)+B(x)]G(x)$. Substituting $x=1$ into this
    and note that $A(1)+B(1)=2n$ and $G(1)ne 0$,
    we obtain $n=2^{k-1}$, as desired.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      I will show that the statement is true with the condition that




      • The two multisets $A={a_i+a_j:1 le i<jle n}$ and
        $B={b_i+b_j:1 le i<jle n}$ are the same
        (i.e. if $x in A$ appears $k$ times in $A$ then
        $x$ appears $k$ times in $B$).


      • Two sets ${a_i}$ and ${b_i}$ are not the same.



      Let $A(x)=sum_{i=1}^n x^{a_i}$ and $B(x)=sum_{i=1}^n x^{b_i}$
      then we have $A^2(x)=A(x^2)+2sum_{iin A} x^i$ and similarly,
      $B^2(x)=B(x^2)+2sum_{iin B} x^i$. Therefore,
      $$A(x^2)-B(x^2)=A^2(x)-B^2(x)=[A(x)+B(x)][A(x)-B(x)].$$
      Since $A(1)=B(1)=n$ so $A(x)-B(x)=(x-1)^kG(x)$ where $G(1)ne 0,
      k ge 1$      .This follows
      $$(x^2-1)^k G(x^2)=A(x^2)-B(x^2)=[A(x)+B(x)]cdot (x-1)^k G(x).$$
      Therefore, $(x+1)^k G(x^2)=[A(x)+B(x)]G(x)$. Substituting $x=1$ into this
      and note that $A(1)+B(1)=2n$ and $G(1)ne 0$,
      we obtain $n=2^{k-1}$, as desired.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        I will show that the statement is true with the condition that




        • The two multisets $A={a_i+a_j:1 le i<jle n}$ and
          $B={b_i+b_j:1 le i<jle n}$ are the same
          (i.e. if $x in A$ appears $k$ times in $A$ then
          $x$ appears $k$ times in $B$).


        • Two sets ${a_i}$ and ${b_i}$ are not the same.



        Let $A(x)=sum_{i=1}^n x^{a_i}$ and $B(x)=sum_{i=1}^n x^{b_i}$
        then we have $A^2(x)=A(x^2)+2sum_{iin A} x^i$ and similarly,
        $B^2(x)=B(x^2)+2sum_{iin B} x^i$. Therefore,
        $$A(x^2)-B(x^2)=A^2(x)-B^2(x)=[A(x)+B(x)][A(x)-B(x)].$$
        Since $A(1)=B(1)=n$ so $A(x)-B(x)=(x-1)^kG(x)$ where $G(1)ne 0,
        k ge 1$        .This follows
        $$(x^2-1)^k G(x^2)=A(x^2)-B(x^2)=[A(x)+B(x)]cdot (x-1)^k G(x).$$
        Therefore, $(x+1)^k G(x^2)=[A(x)+B(x)]G(x)$. Substituting $x=1$ into this
        and note that $A(1)+B(1)=2n$ and $G(1)ne 0$,
        we obtain $n=2^{k-1}$, as desired.






        share|cite|improve this answer









        $endgroup$



        I will show that the statement is true with the condition that




        • The two multisets $A={a_i+a_j:1 le i<jle n}$ and
          $B={b_i+b_j:1 le i<jle n}$ are the same
          (i.e. if $x in A$ appears $k$ times in $A$ then
          $x$ appears $k$ times in $B$).


        • Two sets ${a_i}$ and ${b_i}$ are not the same.



        Let $A(x)=sum_{i=1}^n x^{a_i}$ and $B(x)=sum_{i=1}^n x^{b_i}$
        then we have $A^2(x)=A(x^2)+2sum_{iin A} x^i$ and similarly,
        $B^2(x)=B(x^2)+2sum_{iin B} x^i$. Therefore,
        $$A(x^2)-B(x^2)=A^2(x)-B^2(x)=[A(x)+B(x)][A(x)-B(x)].$$
        Since $A(1)=B(1)=n$ so $A(x)-B(x)=(x-1)^kG(x)$ where $G(1)ne 0,
        k ge 1$        .This follows
        $$(x^2-1)^k G(x^2)=A(x^2)-B(x^2)=[A(x)+B(x)]cdot (x-1)^k G(x).$$
        Therefore, $(x+1)^k G(x^2)=[A(x)+B(x)]G(x)$. Substituting $x=1$ into this
        and note that $A(1)+B(1)=2n$ and $G(1)ne 0$,
        we obtain $n=2^{k-1}$, as desired.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 4:04









        TenguTengu

        2,66411021




        2,66411021























            2












            $begingroup$

            Consider the sequence $(a_k)=(2,4,4)$ and the sequence $(b_k)=(3,3,5)$. Then the two sequences have non-identical sets of elements, but the sets of their pairwise sums are as follows:
            $$
            A={2+4,2+4,4+4}={6,8}
            $$
            $$
            B={3+3,3+5,3+5}={6,8}
            $$



            It would then seem that we need to alter the language to make the statement true... maybe it should be a multiset? Maybe the sequences can't have repeated values?



            EDIT: Some quick and dirty coding indicates the following question rewrite has a better chance of being true:



            Given two sets of natural numbers $A$ and $B$ where $Aneq B$ and $|A|=|B|=n$ such that the restricted sumsets $2^wedge A$ and $2^wedge B$ are equal, show that $n=2^m$ for some natural number $m$.



            To answer @coffeemath's comment in the OP, you can take $A={1,4,5,6}$ and $B={2,3,4,7}$ as examples of the case $n=4$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              RandomMathGuy--- Thanks for the $n=4$example!
              $endgroup$
              – coffeemath
              Dec 10 '18 at 0:14
















            2












            $begingroup$

            Consider the sequence $(a_k)=(2,4,4)$ and the sequence $(b_k)=(3,3,5)$. Then the two sequences have non-identical sets of elements, but the sets of their pairwise sums are as follows:
            $$
            A={2+4,2+4,4+4}={6,8}
            $$
            $$
            B={3+3,3+5,3+5}={6,8}
            $$



            It would then seem that we need to alter the language to make the statement true... maybe it should be a multiset? Maybe the sequences can't have repeated values?



            EDIT: Some quick and dirty coding indicates the following question rewrite has a better chance of being true:



            Given two sets of natural numbers $A$ and $B$ where $Aneq B$ and $|A|=|B|=n$ such that the restricted sumsets $2^wedge A$ and $2^wedge B$ are equal, show that $n=2^m$ for some natural number $m$.



            To answer @coffeemath's comment in the OP, you can take $A={1,4,5,6}$ and $B={2,3,4,7}$ as examples of the case $n=4$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              RandomMathGuy--- Thanks for the $n=4$example!
              $endgroup$
              – coffeemath
              Dec 10 '18 at 0:14














            2












            2








            2





            $begingroup$

            Consider the sequence $(a_k)=(2,4,4)$ and the sequence $(b_k)=(3,3,5)$. Then the two sequences have non-identical sets of elements, but the sets of their pairwise sums are as follows:
            $$
            A={2+4,2+4,4+4}={6,8}
            $$
            $$
            B={3+3,3+5,3+5}={6,8}
            $$



            It would then seem that we need to alter the language to make the statement true... maybe it should be a multiset? Maybe the sequences can't have repeated values?



            EDIT: Some quick and dirty coding indicates the following question rewrite has a better chance of being true:



            Given two sets of natural numbers $A$ and $B$ where $Aneq B$ and $|A|=|B|=n$ such that the restricted sumsets $2^wedge A$ and $2^wedge B$ are equal, show that $n=2^m$ for some natural number $m$.



            To answer @coffeemath's comment in the OP, you can take $A={1,4,5,6}$ and $B={2,3,4,7}$ as examples of the case $n=4$.






            share|cite|improve this answer











            $endgroup$



            Consider the sequence $(a_k)=(2,4,4)$ and the sequence $(b_k)=(3,3,5)$. Then the two sequences have non-identical sets of elements, but the sets of their pairwise sums are as follows:
            $$
            A={2+4,2+4,4+4}={6,8}
            $$
            $$
            B={3+3,3+5,3+5}={6,8}
            $$



            It would then seem that we need to alter the language to make the statement true... maybe it should be a multiset? Maybe the sequences can't have repeated values?



            EDIT: Some quick and dirty coding indicates the following question rewrite has a better chance of being true:



            Given two sets of natural numbers $A$ and $B$ where $Aneq B$ and $|A|=|B|=n$ such that the restricted sumsets $2^wedge A$ and $2^wedge B$ are equal, show that $n=2^m$ for some natural number $m$.



            To answer @coffeemath's comment in the OP, you can take $A={1,4,5,6}$ and $B={2,3,4,7}$ as examples of the case $n=4$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 9 '18 at 3:29

























            answered Dec 9 '18 at 2:44









            RandomMathGuyRandomMathGuy

            212




            212












            • $begingroup$
              RandomMathGuy--- Thanks for the $n=4$example!
              $endgroup$
              – coffeemath
              Dec 10 '18 at 0:14


















            • $begingroup$
              RandomMathGuy--- Thanks for the $n=4$example!
              $endgroup$
              – coffeemath
              Dec 10 '18 at 0:14
















            $begingroup$
            RandomMathGuy--- Thanks for the $n=4$example!
            $endgroup$
            – coffeemath
            Dec 10 '18 at 0:14




            $begingroup$
            RandomMathGuy--- Thanks for the $n=4$example!
            $endgroup$
            – coffeemath
            Dec 10 '18 at 0:14


















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