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Question Number 2820 by Yozzis last updated on 27/Nov/15
We have the idea of Phythagorian triples  as solutions (x,y,z) to the equation                              x^2 +y^2 =z^2   where x,y,z∈Z^+ .   How frequently do solutions (x,y,z,t)  to the equation                                    x^2 +y^2 +z^2 =t^2   arise for x,y,z,t being integers between 1 and 100 inclusively?    What solutions exist for t=12 and x,y,z∈Z^+ ?
$${We}\:{have}\:{the}\:{idea}\:{of}\:{Phythagorian}\:{triples} \\ $$$${as}\:{solutions}\:\left({x},{y},{z}\right)\:{to}\:{the}\:{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={z}^{\mathrm{2}} \\ $$$${where}\:{x},{y},{z}\in\mathbb{Z}^{+} .\: \\ $$$${How}\:{frequently}\:{do}\:{solutions}\:\left({x},{y},{z},{t}\right)\:\:{to}\:{the}\:{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={t}^{\mathrm{2}} \\ $$$${arise}\:{for}\:{x},{y},{z},{t}\:{being}\:{integers}\:{between}\:\mathrm{1}\:{and}\:\mathrm{100}\:{inclusively}? \\ $$$$ \\ $$$${What}\:{solutions}\:{exist}\:{for}\:{t}=\mathrm{12}\:{and}\:{x},{y},{z}\in\mathbb{Z}^{+} ? \\ $$
Commented by prakash jain last updated on 27/Nov/15
Out of 30 number I considered 18 came with  a solution. Some with more than 1.  What exactly are looking for in solution.
$$\mathrm{Out}\:\mathrm{of}\:\mathrm{30}\:\mathrm{number}\:\mathrm{I}\:\mathrm{considered}\:\mathrm{18}\:\mathrm{came}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{solution}.\:\mathrm{Some}\:\mathrm{with}\:\mathrm{more}\:\mathrm{than}\:\mathrm{1}. \\ $$$$\mathrm{What}\:\mathrm{exactly}\:\mathrm{are}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{in}\:\mathrm{solution}. \\ $$
Commented by Yozzis last updated on 28/Nov/15
I was thinking sonething like the prime  density function ((π(N))/N) but instead  call something the solution density function ((η(N))/N)  where η(N) is the number of solutions (x,y,z,t)  for 1≤x,y,z,t≤N , N∈N.   So, if N=2 for example, η(2)=0 since  there are is solution (x,y,z,t) with x,y,z,t  each taking either 1 or 2 only.  So,((η(N))/N)=(0/2)=0.   In my question N=100.
$${I}\:{was}\:{thinking}\:{sonething}\:{like}\:{the}\:{prime} \\ $$$${density}\:{function}\:\frac{\pi\left({N}\right)}{{N}}\:{but}\:{instead} \\ $$$${call}\:{something}\:{the}\:{solution}\:{density}\:{function}\:\frac{\eta\left({N}\right)}{{N}} \\ $$$${where}\:\eta\left({N}\right)\:{is}\:{the}\:{number}\:{of}\:{solutions}\:\left({x},{y},{z},{t}\right) \\ $$$${for}\:\mathrm{1}\leqslant{x},{y},{z},{t}\leqslant{N}\:,\:{N}\in\mathbb{N}.\: \\ $$$${So},\:{if}\:{N}=\mathrm{2}\:{for}\:{example},\:\eta\left(\mathrm{2}\right)=\mathrm{0}\:{since} \\ $$$${there}\:{are}\:{is}\:{solution}\:\left({x},{y},{z},{t}\right)\:{with}\:{x},{y},{z},{t} \\ $$$${each}\:{taking}\:{either}\:\mathrm{1}\:{or}\:\mathrm{2}\:{only}. \\ $$$${So},\frac{\eta\left({N}\right)}{{N}}=\frac{\mathrm{0}}{\mathrm{2}}=\mathrm{0}.\: \\ $$$${In}\:{my}\:{question}\:{N}=\mathrm{100}. \\ $$
Commented by Yozzis last updated on 27/Nov/15
If we consider distinct solutions ((η(N))/N) is better defined  I think.
$${If}\:{we}\:{consider}\:{distinct}\:{solutions}\:\frac{\eta\left({N}\right)}{{N}}\:{is}\:{better}\:{defined} \\ $$$${I}\:{think}. \\ $$
Commented by prakash jain last updated on 27/Nov/15
I think you should define solution density  as ((η(N))/N), where t≤N. Since you are trying  to find how many of them can be written  as sum of three squares.  I mean no need to impose restriction on  x,y,z.  If you want to define it how many such  quaruplets can be found then considering  distinct solution is correct.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{should}\:\mathrm{define}\:\mathrm{solution}\:\mathrm{density} \\ $$$$\mathrm{as}\:\frac{\eta\left(\mathrm{N}\right)}{\mathrm{N}},\:\mathrm{where}\:{t}\leqslant\mathrm{N}.\:\mathrm{Since}\:\mathrm{you}\:\mathrm{are}\:\mathrm{trying} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{how}\:\mathrm{many}\:\mathrm{of}\:\mathrm{them}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written} \\ $$$$\mathrm{as}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{three}\:\mathrm{squares}. \\ $$$$\mathrm{I}\:\mathrm{mean}\:\mathrm{no}\:\mathrm{need}\:\mathrm{to}\:\mathrm{impose}\:\mathrm{restriction}\:\mathrm{on} \\ $$$${x},{y},{z}. \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{define}\:\mathrm{it}\:\mathrm{how}\:\mathrm{many}\:\mathrm{such} \\ $$$$\mathrm{quaruplets}\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}\:\mathrm{then}\:\mathrm{considering} \\ $$$$\mathrm{distinct}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{correct}. \\ $$
Commented by Yozzi last updated on 27/Nov/15
Makes sense.
$${Makes}\:{sense}.\: \\ $$
Commented by prakash jain last updated on 27/Nov/15
It is easy to see that if x,y,z,t is solution  than kx,ky,kz,kt also a solution.  Which makes desnity likely to be higher  for higher N.
$$\mathrm{It}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{that}\:\mathrm{if}\:{x},{y},{z},{t}\:\mathrm{is}\:\mathrm{solution} \\ $$$$\mathrm{than}\:{kx},{ky},{kz},{kt}\:\mathrm{also}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$\mathrm{Which}\:\mathrm{makes}\:\mathrm{desnity}\:\mathrm{likely}\:\mathrm{to}\:\mathrm{be}\:\mathrm{higher} \\ $$$$\mathrm{for}\:\mathrm{higher}\:\mathrm{N}. \\ $$
Commented by prakash jain last updated on 27/Nov/15
What is the density function for  Pythogorian triplets.
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{density}\:\mathrm{function}\:\mathrm{for} \\ $$$$\mathrm{Pythogorian}\:\mathrm{triplets}. \\ $$
Commented by Yozzis last updated on 28/Nov/15
I don′t even know if solution  density functions is a thing  hahaha...  It may have a different name.  I don′t know anything much about number theory.  Maybe Google could help.
$${I}\:{don}'{t}\:{even}\:{know}\:{if}\:{solution} \\ $$$${density}\:{functions}\:{is}\:{a}\:{thing}\:\:\mathrm{hahaha}… \\ $$$${It}\:{may}\:{have}\:{a}\:{different}\:{name}. \\ $$$${I}\:{don}'{t}\:{know}\:{anything}\:{much}\:{about}\:{number}\:{theory}. \\ $$$${Maybe}\:{Google}\:{could}\:{help}. \\ $$
Commented by prakash jain last updated on 27/Nov/15
I didn′t hear about it either. Was wondering  why you didn′t start with finding density  function for pythgorian triplets.
$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{hear}\:\mathrm{about}\:\mathrm{it}\:\mathrm{either}.\:\mathrm{Was}\:\mathrm{wondering} \\ $$$$\mathrm{why}\:\mathrm{you}\:\mathrm{didn}'\mathrm{t}\:\mathrm{start}\:\mathrm{with}\:\mathrm{finding}\:\mathrm{density} \\ $$$$\mathrm{function}\:\mathrm{for}\:\mathrm{pythgorian}\:\mathrm{triplets}. \\ $$
Commented by Yozzis last updated on 27/Nov/15
I wouldn′t know where to start :(.
$${I}\:{wouldn}'{t}\:{know}\:{where}\:{to}\:{start}\::\left(.\right. \\ $$
Commented by prakash jain last updated on 28/Nov/15
I asked this questiin since you are sort of  extending the definition.  So it would probably be good to go thru the  work already done on pythogorean triples.  This may help and coming up with a good  definition of  ((η(N))/N) and also maybe help  in generalizing to Σ_(i=1) ^n x_i ^2 =t^2 .  As shown in my part answer we only need to consider  only odd t.  I will add a new question for discussion.
$$\mathrm{I}\:\mathrm{asked}\:\mathrm{this}\:\mathrm{questiin}\:\mathrm{since}\:\mathrm{you}\:\mathrm{are}\:\mathrm{sort}\:\mathrm{of} \\ $$$$\mathrm{extending}\:\mathrm{the}\:\mathrm{definition}. \\ $$$$\mathrm{So}\:\mathrm{it}\:\mathrm{would}\:\mathrm{probably}\:\mathrm{be}\:\mathrm{good}\:\mathrm{to}\:\mathrm{go}\:\mathrm{thru}\:\mathrm{the} \\ $$$$\mathrm{work}\:\mathrm{already}\:\mathrm{done}\:\mathrm{on}\:\mathrm{pythogorean}\:\mathrm{triples}. \\ $$$$\mathrm{This}\:\mathrm{may}\:\mathrm{help}\:\mathrm{and}\:\mathrm{coming}\:\mathrm{up}\:\mathrm{with}\:\mathrm{a}\:\mathrm{good} \\ $$$$\mathrm{definition}\:\mathrm{of}\:\:\frac{\eta\left({N}\right)}{{N}}\:\mathrm{and}\:\mathrm{also}\:\mathrm{maybe}\:\mathrm{help} \\ $$$$\mathrm{in}\:\mathrm{generalizing}\:\mathrm{to}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{i}} ^{\mathrm{2}} ={t}^{\mathrm{2}} . \\ $$$$\mathrm{As}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{my}\:\mathrm{part}\:\mathrm{answer}\:\mathrm{we}\:\mathrm{only}\:\mathrm{need}\:\mathrm{to}\:\mathrm{consider} \\ $$$$\mathrm{only}\:\mathrm{odd}\:{t}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{add}\:\mathrm{a}\:\mathrm{new}\:\mathrm{question}\:\mathrm{for}\:\mathrm{discussion}. \\ $$
Answered by prakash jain last updated on 27/Nov/15
PART A: t even     t=2n     A.1: x odd, y odd, z even     (2j+1)^2 +(2k+1)^2 +(2l)^2 =4n^2      4j^2 +4j+1+4k^2 +4k+1+4l^2 =4n^2      ⇒n^2 =j^2 +j+k^2 +k+l^2 +(1/2) (fraction)      A.1 no solution with 2 odd number.        A.2: x,y,z even      4j^2 +4k^2 +4l^2 =4n^2       j^2 +k^2 +l^2 =n^2       same condition for j,k,l  From above we can conclude that if  t=2^m ∙n (where n is odd number) then  x=2^m .i, y=2^m .j, z=2^m .k  Solving for t=12  t=2^2 .3, x=2^2 i, y=2^2 .j,z=2^2 k  i^2 +j^2 +k^2 =9  Only only solution (excluding rearrangements)  i=2,j=2,k=1  x=8,y=8,z=4  Will consider t odd cases in seperate post.
$$\mathrm{PART}\:\mathrm{A}:\:{t}\:{even} \\ $$$$\:\:\:{t}=\mathrm{2}{n} \\ $$$$\:\:\:{A}.\mathrm{1}:\:{x}\:{odd},\:{y}\:{odd},\:{z}\:{even} \\ $$$$\:\:\:\left(\mathrm{2}{j}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{l}\right)^{\mathrm{2}} =\mathrm{4}{n}^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{4}{j}^{\mathrm{2}} +\mathrm{4}{j}+\mathrm{1}+\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1}+\mathrm{4}{l}^{\mathrm{2}} =\mathrm{4}{n}^{\mathrm{2}} \\ $$$$\:\:\:\Rightarrow{n}^{\mathrm{2}} ={j}^{\mathrm{2}} +{j}+{k}^{\mathrm{2}} +{k}+{l}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{fraction}\right) \\ $$$$\:\:\:\:{A}.\mathrm{1}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{2}\:\mathrm{odd}\:\mathrm{number}. \\ $$$$ \\ $$$$\:\:\:\:{A}.\mathrm{2}:\:{x},{y},{z}\:{even} \\ $$$$\:\:\:\:\mathrm{4}{j}^{\mathrm{2}} +\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{l}^{\mathrm{2}} =\mathrm{4}{n}^{\mathrm{2}} \\ $$$$\:\:\:\:{j}^{\mathrm{2}} +{k}^{\mathrm{2}} +{l}^{\mathrm{2}} ={n}^{\mathrm{2}} \\ $$$$\:\:\:\:{same}\:{condition}\:{for}\:{j},{k},{l} \\ $$$$\mathrm{From}\:\mathrm{above}\:\mathrm{we}\:\mathrm{can}\:\mathrm{conclude}\:\mathrm{that}\:\mathrm{if} \\ $$$${t}=\mathrm{2}^{{m}} \centerdot{n}\:\left({where}\:{n}\:{is}\:{odd}\:{number}\right)\:{then} \\ $$$${x}=\mathrm{2}^{{m}} .{i},\:{y}=\mathrm{2}^{{m}} .{j},\:{z}=\mathrm{2}^{{m}} .{k} \\ $$$$\boldsymbol{\mathrm{Solving}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{t}}=\mathrm{12} \\ $$$${t}=\mathrm{2}^{\mathrm{2}} .\mathrm{3},\:{x}=\mathrm{2}^{\mathrm{2}} {i},\:{y}=\mathrm{2}^{\mathrm{2}} .{j},{z}=\mathrm{2}^{\mathrm{2}} {k} \\ $$$${i}^{\mathrm{2}} +{j}^{\mathrm{2}} +{k}^{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{Only}\:\mathrm{only}\:\mathrm{solution}\:\left({excluding}\:{rearrangements}\right) \\ $$$${i}=\mathrm{2},{j}=\mathrm{2},{k}=\mathrm{1} \\ $$$${x}=\mathrm{8},{y}=\mathrm{8},{z}=\mathrm{4} \\ $$$$\mathrm{Will}\:\mathrm{consider}\:{t}\:\mathrm{odd}\:\mathrm{cases}\:\mathrm{in}\:\mathrm{seperate}\:\mathrm{post}. \\ $$
Answered by prakash jain last updated on 27/Nov/15
PART B: t odd  t=2n+1       case A: x, y, z odd       (2l+1)^2 +(2m+1)^2 +(2k+1)^2 =(2n+1)^2         4l^2 +4l+1+4m^2 +4m+1+4k^2 +4k+1=4n^2 +4n+1        n^2 =(integer)+(1/2)        so no solution for x,y,z and t odd.         case B: x,y even z odd       4j^2 +4k^2 +(2l−1)^2 =4n^2 +4n+1       j^2 +k^2 +l^2 −l=n^2 +n       solvable.  I will do more work to see what further  condition can be added to solve these cases.
$$\mathrm{PART}\:\mathrm{B}:\:{t}\:{odd} \\ $$$${t}=\mathrm{2}{n}+\mathrm{1} \\ $$$$\:\:\:\:\:{case}\:{A}:\:{x},\:{y},\:{z}\:{odd} \\ $$$$\:\:\:\:\:\left(\mathrm{2}{l}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{4}{l}^{\mathrm{2}} +\mathrm{4}{l}+\mathrm{1}+\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}{m}+\mathrm{1}+\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1}=\mathrm{4}{n}^{\mathrm{2}} +\mathrm{4}{n}+\mathrm{1} \\ $$$$\:\:\:\:\:\:{n}^{\mathrm{2}} =\left({integer}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{so}\:{no}\:{solution}\:{for}\:{x},{y},{z}\:{and}\:{t}\:{odd}. \\ $$$$ \\ $$$$\:\:\:\:\:{case}\:{B}:\:{x},{y}\:{even}\:{z}\:{odd} \\ $$$$\:\:\:\:\:\mathrm{4}{j}^{\mathrm{2}} +\mathrm{4}{k}^{\mathrm{2}} +\left(\mathrm{2}{l}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{n}^{\mathrm{2}} +\mathrm{4}{n}+\mathrm{1} \\ $$$$\:\:\:\:\:{j}^{\mathrm{2}} +{k}^{\mathrm{2}} +{l}^{\mathrm{2}} −{l}={n}^{\mathrm{2}} +{n} \\ $$$$\:\:\:\:\:{solvable}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{do}\:\mathrm{more}\:\mathrm{work}\:\mathrm{to}\:\mathrm{see}\:\mathrm{what}\:\mathrm{further} \\ $$$$\mathrm{condition}\:\mathrm{can}\:\mathrm{be}\:\mathrm{added}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{these}\:\mathrm{cases}. \\ $$
Answered by prakash jain last updated on 27/Nov/15
Some Solution for t till 100  smallest solution start t=3  t=2^k =2,4,8,16,32,64 (no solutions)  t=2^k ∙3=3,6,12,24,48,96 x=2^k .2,y=2^k .2,z=2^k   t=2^k .5=5,10,20,40,80 (no solutions)  t=2^k ∙7=7,14,28,56  x=2^k ∙6, y=2^k ∙3, z=2^k .2  t=2^k .9=9,18,36,72  (x,y,z)=(8,4,1),(7,4,4),(6,3,3)  t=2^k ∙11=11,22,44,88 (x,y,z)=(9,6,2),(7,6,6)  I think I may not be answering the question  the way you want.
$$\mathrm{Some}\:\mathrm{Solution}\:\mathrm{for}\:{t}\:{till}\:\mathrm{100} \\ $$$${smallest}\:{solution}\:{start}\:{t}=\mathrm{3} \\ $$$${t}=\mathrm{2}^{{k}} =\mathrm{2},\mathrm{4},\mathrm{8},\mathrm{16},\mathrm{32},\mathrm{64}\:\left(\mathrm{no}\:\mathrm{solutions}\right) \\ $$$${t}=\mathrm{2}^{{k}} \centerdot\mathrm{3}=\mathrm{3},\mathrm{6},\mathrm{12},\mathrm{24},\mathrm{48},\mathrm{96}\:{x}=\mathrm{2}^{{k}} .\mathrm{2},{y}=\mathrm{2}^{{k}} .\mathrm{2},{z}=\mathrm{2}^{{k}} \\ $$$${t}=\mathrm{2}^{{k}} .\mathrm{5}=\mathrm{5},\mathrm{10},\mathrm{20},\mathrm{40},\mathrm{80}\:\left({no}\:{solutions}\right) \\ $$$${t}=\mathrm{2}^{{k}} \centerdot\mathrm{7}=\mathrm{7},\mathrm{14},\mathrm{28},\mathrm{56}\:\:{x}=\mathrm{2}^{{k}} \centerdot\mathrm{6},\:{y}=\mathrm{2}^{{k}} \centerdot\mathrm{3},\:{z}=\mathrm{2}^{{k}} .\mathrm{2} \\ $$$${t}=\mathrm{2}^{{k}} .\mathrm{9}=\mathrm{9},\mathrm{18},\mathrm{36},\mathrm{72}\:\:\left({x},{y},{z}\right)=\left(\mathrm{8},\mathrm{4},\mathrm{1}\right),\left(\mathrm{7},\mathrm{4},\mathrm{4}\right),\left(\mathrm{6},\mathrm{3},\mathrm{3}\right) \\ $$$${t}=\mathrm{2}^{{k}} \centerdot\mathrm{11}=\mathrm{11},\mathrm{22},\mathrm{44},\mathrm{88}\:\left({x},{y},{z}\right)=\left(\mathrm{9},\mathrm{6},\mathrm{2}\right),\left(\mathrm{7},\mathrm{6},\mathrm{6}\right) \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{may}\:{not}\:\mathrm{be}\:\mathrm{answering}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{the}\:\mathrm{way}\:\mathrm{you}\:\mathrm{want}. \\ $$

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