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Question Number 2820 by Yozzis last updated on 27/Nov/15

W e h a v e t h e i d e a o f P h y t h a g o r i a n t r i p l e s

a s s o l u t i o n s (x, y, z) t o t h e e q u a t i o n

x^{2} + y^{2} = z^{2}

w h e r e x, y, z \in Z^{+} .

H o w f r e q u e n t l y d o s o l u t i o n s (x, y, z, t) t o t h e e q u a t i o n

x^{2} + y^{2} + z^{2} = t^{2}

a r i s e f o r x, y, z, t b e i n g i n t e g e r s b e t w e e n 1 a n d 100 i n c l u s i v e l y ?

W h a t s o l u t i o n s e x i s t f o r t = 12 a n d x, y, z \in 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 .

Commented by Yozzis last updated on 28/Nov/15

I w a s t h i n k i n g s o n e t h i n g l i k e t h e p r i m e

d e n s i t y f u n c t i o n \frac{π (N)}{N} b u t i n s t e a d

c a l l s o m e t h i n g t h e s o l u t i o n d e n s i t y f u n c t i o n \frac{η (N)}{N}

w h e r e η (N) i s t h e n u m b e r o f s o l u t i o n s (x, y, z, t)

f o r 1 ⩽ x, y, z, t ⩽ N, N \in N .

S o, i f N = 2 f o r e x a m p l e, η (2) = 0 s i n c e

t h e r e a r e i s s o l u t i o n (x, y, z, t) w i t h x, y, z, t

e a c h t a k i n g e i t h e r 1 o r 2 o n l y .

S o, \frac{η (N)}{N} = \frac{0}{2} = 0 .

I n m y q u e s t i o n N = 100 .

Commented by Yozzis last updated on 27/Nov/15

I f w e c o n s i d e r d i s t i n c t s o l u t i o n s \frac{η (N)}{N} i s b e t t e r d e f i n e d

I t h i n k .

Commented by prakash jain last updated on 27/Nov/15

I think you should define solution density

as \frac{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 .

Commented by Yozzi last updated on 27/Nov/15

M a k e s s e n s e .

Commented by prakash jain last updated on 27/Nov/15

It is easy to see that if x, y, z, t is solution

than k x, k y, k z, k t also a solution .

Which makes desnity likely to be higher

for higher N .

Commented by prakash jain last updated on 27/Nov/15

What is the density function for

Pythogorian triplets .

Commented by Yozzis last updated on 28/Nov/15

I {d o n}^{'} t e v e n k n o w i f s o l u t i o n

d e n s i t y f u n c t i o n s i s a t h i n g hahaha \dots

I t m a y h a v e a d i f f e r e n t n a m e .

I {d o n}^{'} t k n o w a n y t h i n g m u c h a b o u t n u m b e r t h e o r y .

M a y b e G o o g l e c o u l d h e l p .

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 .

Commented by Yozzis last updated on 27/Nov/15

I {w o u l d n}^{'} t k n o w w h e r e t o s t a r t : (.

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 \frac{η (N)}{N} and also maybe help

in generalizing to \underset{i = 1}{\sum^{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 .

Answered by prakash jain last updated on 27/Nov/15

PART A : t e v e n

t = 2 n

A . 1 : x o d d, y o d d, z e v e n

{(2 j + 1)}^{2} + {(2 k + 1)}^{2} + {(2 l)}^{2} = 4 n^{2}

4 j^{2} + 4 j + 1 + 4 k^{2} + 4 k + 1 + 4 l^{2} = 4 n^{2}

\Rightarrow n^{2} = j^{2} + j + k^{2} + k + l^{2} + \frac{1}{2} (fraction)

A . 1 no solution with 2 odd number .

A . 2 : x, y, z e v e n

4 j^{2} + 4 k^{2} + 4 l^{2} = 4 n^{2}

j^{2} + k^{2} + l^{2} = n^{2}

s a m e c o n d i t i o n f o r j, k, l

From above we can conclude that if

t = 2^{m} \cdot n (w h e r e n i s o d d n u m b e r) t h e n

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 (e x c l u d i n g r e a r r a n g e m e n t s)

i = 2, j = 2, k = 1

x = 8, y = 8, z = 4

Will consider t odd cases in seperate post .

Answered by prakash jain last updated on 27/Nov/15

PART B : t o d d

t = 2 n + 1

c a s e A : x, y, z o d d

{(2 l + 1)}^{2} + {(2 m + 1)}^{2} + {(2 k + 1)}^{2} = {(2 n + 1)}^{2}

4 l^{2} + 4 l + 1 + 4 m^{2} + 4 m + 1 + 4 k^{2} + 4 k + 1 = 4 n^{2} + 4 n + 1

n^{2} = (i n t e g e r) + \frac{1}{2}

s o n o s o l u t i o n f o r x, y, z a n d t o d d .

c a s e B : x, y e v e n z o d d

4 j^{2} + 4 k^{2} + {(2 l - 1)}^{2} = 4 n^{2} + 4 n + 1

j^{2} + k^{2} + l^{2} - l = n^{2} + n

s o l v a b l e .

I will do more work to see what further

condition can be added to solve these cases .

Answered by prakash jain last updated on 27/Nov/15

Some Solution for t t i l l 100

s m a l l e s t s o l u t i o n s t a r t t = 3

t = 2^{k} = 2, 4, 8, 16, 32, 64 (no solutions)

t = 2^{k} \cdot 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 (n o s o l u t i o n s)

t = 2^{k} \cdot 7 = 7, 14, 28, 56 x = 2^{k} \cdot 6, y = 2^{k} \cdot 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} \cdot 11 = 11, 22, 44, 88 (x, y, z) = (9, 6, 2), (7, 6, 6)

I think I may n o t be answering the question

the way you want .