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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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