Pyramid Of Spheres












1












$begingroup$


A collection of identical spheres can be formed into a “square” pyramid (a pyramid with a base (bottom layer) made up of $n times n$ spheres whose next layer is made up of $(n-1) times (n-1)$ spheres, continuing this way up to the top layer of one sphere). The same collection of spheres can also be formed into a single-layer $k times k$ “square” where $k < 100$.




Find the largest possible value of $k$ for such a collection of spheres.




It seems that the answer will be the square of a number less than $100$ and the approach I took was saying
$$
k^2= 1+ 2^2+ ....(n-1)^2 + n^2
$$

since each "level" of the pyramid has exactly $n^2$ spheres for any given level $n$ of the pyramid. The sum of these layers or levels will be $k^2$
but how this sum can be evaluated is where Me stuck :(










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
    $endgroup$
    – Will Jagy
    Dec 3 '18 at 20:17












  • $begingroup$
    W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
    $endgroup$
    – Will Jagy
    Dec 3 '18 at 20:30










  • $begingroup$
    will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
    $endgroup$
    – Randin
    Dec 3 '18 at 20:42
















1












$begingroup$


A collection of identical spheres can be formed into a “square” pyramid (a pyramid with a base (bottom layer) made up of $n times n$ spheres whose next layer is made up of $(n-1) times (n-1)$ spheres, continuing this way up to the top layer of one sphere). The same collection of spheres can also be formed into a single-layer $k times k$ “square” where $k < 100$.




Find the largest possible value of $k$ for such a collection of spheres.




It seems that the answer will be the square of a number less than $100$ and the approach I took was saying
$$
k^2= 1+ 2^2+ ....(n-1)^2 + n^2
$$

since each "level" of the pyramid has exactly $n^2$ spheres for any given level $n$ of the pyramid. The sum of these layers or levels will be $k^2$
but how this sum can be evaluated is where Me stuck :(










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
    $endgroup$
    – Will Jagy
    Dec 3 '18 at 20:17












  • $begingroup$
    W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
    $endgroup$
    – Will Jagy
    Dec 3 '18 at 20:30










  • $begingroup$
    will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
    $endgroup$
    – Randin
    Dec 3 '18 at 20:42














1












1








1





$begingroup$


A collection of identical spheres can be formed into a “square” pyramid (a pyramid with a base (bottom layer) made up of $n times n$ spheres whose next layer is made up of $(n-1) times (n-1)$ spheres, continuing this way up to the top layer of one sphere). The same collection of spheres can also be formed into a single-layer $k times k$ “square” where $k < 100$.




Find the largest possible value of $k$ for such a collection of spheres.




It seems that the answer will be the square of a number less than $100$ and the approach I took was saying
$$
k^2= 1+ 2^2+ ....(n-1)^2 + n^2
$$

since each "level" of the pyramid has exactly $n^2$ spheres for any given level $n$ of the pyramid. The sum of these layers or levels will be $k^2$
but how this sum can be evaluated is where Me stuck :(










share|cite|improve this question











$endgroup$




A collection of identical spheres can be formed into a “square” pyramid (a pyramid with a base (bottom layer) made up of $n times n$ spheres whose next layer is made up of $(n-1) times (n-1)$ spheres, continuing this way up to the top layer of one sphere). The same collection of spheres can also be formed into a single-layer $k times k$ “square” where $k < 100$.




Find the largest possible value of $k$ for such a collection of spheres.




It seems that the answer will be the square of a number less than $100$ and the approach I took was saying
$$
k^2= 1+ 2^2+ ....(n-1)^2 + n^2
$$

since each "level" of the pyramid has exactly $n^2$ spheres for any given level $n$ of the pyramid. The sum of these layers or levels will be $k^2$
but how this sum can be evaluated is where Me stuck :(







combinatorics word-problem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 20:48









Daniele Tampieri

2,1621621




2,1621621










asked Dec 3 '18 at 20:11









RandinRandin

347116




347116








  • 2




    $begingroup$
    as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
    $endgroup$
    – Will Jagy
    Dec 3 '18 at 20:17












  • $begingroup$
    W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
    $endgroup$
    – Will Jagy
    Dec 3 '18 at 20:30










  • $begingroup$
    will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
    $endgroup$
    – Randin
    Dec 3 '18 at 20:42














  • 2




    $begingroup$
    as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
    $endgroup$
    – Will Jagy
    Dec 3 '18 at 20:17












  • $begingroup$
    W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
    $endgroup$
    – Will Jagy
    Dec 3 '18 at 20:30










  • $begingroup$
    will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
    $endgroup$
    – Randin
    Dec 3 '18 at 20:42








2




2




$begingroup$
as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
$endgroup$
– Will Jagy
Dec 3 '18 at 20:17






$begingroup$
as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
$endgroup$
– Will Jagy
Dec 3 '18 at 20:17














$begingroup$
W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
$endgroup$
– Will Jagy
Dec 3 '18 at 20:30




$begingroup$
W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
$endgroup$
– Will Jagy
Dec 3 '18 at 20:30












$begingroup$
will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
$endgroup$
– Randin
Dec 3 '18 at 20:42




$begingroup$
will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
$endgroup$
– Randin
Dec 3 '18 at 20:42










1 Answer
1






active

oldest

votes


















2












$begingroup$

If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.



Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.



Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.



We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.



If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024596%2fpyramid-of-spheres%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.



    Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.



    Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.



    We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.



    If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.



      Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.



      Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.



      We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.



      If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.



        Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.



        Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.



        We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.



        If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.






        share|cite|improve this answer











        $endgroup$



        If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.



        Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.



        Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.



        We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.



        If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 21:46

























        answered Dec 3 '18 at 21:30









        Oscar LanziOscar Lanzi

        12.4k12036




        12.4k12036






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024596%2fpyramid-of-spheres%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Plaza Victoria

            Puebla de Zaragoza

            Musa