Let-n-j-q-Z-1-Are-there-triples-n-j-q-such-that-the-following-conditions-are-satisfied-altogether-i-n-j-q-ii-n-2-j-2-q-2-Suppose-then-that-condition-ii-

Question Number 5144 by Yozzii last updated on 19/Apr/16

L e t n, j, q \in (Z^{+} - {1}) . A r e t h e r e

t r i p l e s (n, j, q) s u c h t h a t t h e f o l l o w i n g

c o n d i t i o n s a r e s a t i s f i e d a l t o g e t h e r ?

(i) n = j^{q}

(i i) n^{2} = j^{2} + q^{2}

- - - - - - - - - - - - - - - - - - - - - -

S u p p o s e t h e n t h a t c o n d i t i o n (i i) a b o v e

i s r e p l a c e d b y t h e f o l l o w i n g c o n d i t i o n :

(i i i) n^{2} = {r j}^{2} + q^{2} w h e r e r \in (Z - {0, 1})

W h a t s o l u t i o n s (n, j, q) e x i s t i n t h i s c a s e ?

Commented by prakash jain last updated on 19/Apr/16

Let us n is even

n = 2 a

c a s e a : j a n d q o d d

4 a^{2} = {(2 b + 1)}^{2} + {(2 c + 1)}^{2} = 4 (b^{2} + c^{2} + b + c) + 2

i m p o s s i b l e

s o b o t h j a n d q m u s t b e e v e n .

c o n t i n u i n g t h e s a m e a r g u m e n t

i f n = 2^{k} a (a i s o d d) (k ⩾ 1)

t h e n q = 2^{k} b, j = 2^{k} c

2^{k} a = {(2^{k} c)}^{2 k b}

2^{k} a = {(2^{k})}^{2 k b} c^{2 k b}

a = {(2^{k})}^{2 k b - 1} c^{2 k b}

i m p o s s i b l e s i n c e a i s o d d n u m b e r

s o n o e v e n s o l u t i o n s f o r n f o r c a s e (i)

c o n t i n u e f o r c a s e s w h e r e n i s o d d

Commented by prakash jain last updated on 20/Apr/16

So no solution for n is possible for which

i f n i s o d d j m u s t b e o d d a n d q e v e n

s i n c e n = j^{q}

a l s o

n^{2} = j^{2} + q^{2}

j^{2 q} = j^{2} + q^{2}

j^{2} (j^{2 q - 2} - 1) = q^{2}

\Rightarrow {(\frac{q}{j})}^{2} = j^{2 q - 2} - 1

s i n c e j i s o d d a n d q e v e n t h e a b o v e r e l a t i o n

i s i m p o s s i b l e .

c o r r e c t i o n : the above argument that q e v r n

a n d j o d d i s n o t c o r r e c t . So the above

relation may be possible .

So no solution for n is possible for which

satisfied both (i) a n d (i i) .

Commented by prakash jain last updated on 19/Apr/16

f o r c o n d i t i o n (i i i) d o y o u m e a n r, n, j, q c a n

t a k e - v e v a l u e s a s w e l l .

Commented by Yozzii last updated on 19/Apr/16

Y e s, b u t o n l y c a n r b e n e g a t i v e

(o r I o r i g i n a l l y p o s t e d f o r r t o p o s s i b l y

b e n e g a t i v e) .

Commented by prakash jain last updated on 19/Apr/16

j = 2^{b}

n = 2^{b q}

r = \frac{n^{2} - q^{2}}{j^{2}} = \frac{2^{b q} - q^{2}}{2^{2 b}}

choose q = 2^{b} \cdot k

r w i l l b e a n i n t e g e r

e x .

j = 2^{2} = 4

q = 2^{2} \cdot k s a y = 2^{2} \cdot 3 = 12

n = 4^{12}

r = \frac{4^{24} - 12^{2}}{4^{2}} = i n t e g e r

So infinite solution can be found .

Given any j we can always choose q such that

r is an integer .

n^{2} > q^{2} so r will always be + ve .