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) above is replaced by the following condition: (iii) n^2 =rj^2 +q^2 where r∈(Z−{0,1}) What solutions (n,j,q) exist in this case?


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Question Number 5144 by Yozzii last updated on 19/Apr/16
$Let n,j,q∈(Z^+ −{1}). Are there triples (n,j,q) such that the following conditions are satisfied altogether? (i) n=j^q (ii)n^2 =j^2 +q^2 −−−−−−−−−−−−−−−−−−−−−− Suppose then that condition (ii) above is replaced by the following condition: (iii) n^2 =rj^2 +q^2 where r∈(Z−{0,1}) What solutions (n,j,q) exist in this case?$
$L e t n, j, q \in (Z^{+} - {1}) . A r e t h e r e$ $t r i p l e s (n, j, q) s u c h t h a t t h e f o l l o w i n g$ $c o n d i t i o n s a r e s a t i s f i e d a l t o g e t h e r ?$ $(i) n = j^{q}$ $(i i) n^{2} = j^{2} + q^{2}$ $- - - - - - - - - - - - - - - - - - - - - -$ $S u p p o s e t h e n t h a t c o n d i t i o n (i i) a b o v e$ $i s r e p l a c e d b y t h e f o l l o w i n g c o n d i t i o n :$ $(i i i) n^{2} = {r j}^{2} + q^{2} w h e r e r \in (Z - {0, 1})$ $W h a t s o l u t i o n s (n, j, q) e x i s t i n t h i s c a s e ?$
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

$j = 2^{b}$ $n = 2^{b q}$ $r = \frac{n^{2} - q^{2}}{j^{2}} = \frac{2^{b q} - q^{2}}{2^{2 b}}$ $choose q = 2^{b} \cdot k$ $r w i l l b e a n i n t e g e r$ $e x .$ $j = 2^{2} = 4$ $q = 2^{2} \cdot k s a y = 2^{2} \cdot 3 = 12$ $n = 4^{12}$ $r = \frac{4^{24} - 12^{2}}{4^{2}} = i n t e g e r$ $So infinite solution can be found .$ $Given any j we can always choose q such that$ $r is an integer .$ $n^{2} > q^{2} so r will always be + ve .$
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