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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?

$${Let}\:{n},{j},{q}\in\left(\mathbb{Z}^{+} −\left\{\mathrm{1}\right\}\right).\:{Are}\:{there}\: \\ $$$${triples}\:\left({n},{j},{q}\right)\:{such}\:{that}\:{the}\:{following} \\ $$$${conditions}\:{are}\:{satisfied}\:{altogether}? \\ $$$$\left({i}\right)\:{n}={j}^{{q}} \:\:\:\: \\ $$$$\left({ii}\right){n}^{\mathrm{2}} ={j}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Suppose}\:{then}\:{that}\:{condition}\:\left({ii}\right)\:{above} \\ $$$${is}\:{replaced}\:{by}\:{the}\:{following}\:{condition}: \\ $$$$ \\ $$$$\left({iii}\right)\:{n}^{\mathrm{2}} ={rj}^{\mathrm{2}} +{q}^{\mathrm{2}} \:{where}\:{r}\in\left(\mathbb{Z}−\left\{\mathrm{0},\mathrm{1}\right\}\right) \\ $$$$ \\ $$$${What}\:{solutions}\:\left({n},{j},{q}\right)\:{exist}\:{in}\:{this}\:{case}? \\ $$$$ \\ $$

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

Let us n is even n=2a case a: j and q odd 4a^2 =(2b+1)^2 +(2c+1)^2 =4(b^2 +c^2 +b+c)+2 impossible so both j and q must be even. continuing the same argument if n=2^k a (a is odd) (k≥1) then q=2^k b, j=2^k c 2^k a=(2^k c)^(2kb) 2^k a=(2^k )^(2kb) c^(2kb) a=(2^k )^(2kb−1) c^(2kb) impossible since a is odd number so no even solutions for n for case (i) continue for cases where n is odd

$$\mathrm{Let}\:\mathrm{us}\:{n}\:\mathrm{is}\:\mathrm{even} \\ $$$${n}=\mathrm{2}{a} \\ $$$${case}\:{a}:\:{j}\:{and}\:{q}\:{odd} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} =\left(\mathrm{2}{b}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{c}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{b}+{c}\right)+\mathrm{2} \\ $$$${impossible} \\ $$$${so}\:{both}\:{j}\:{and}\:{q}\:{must}\:{be}\:{even}. \\ $$$${continuing}\:{the}\:{same}\:{argument}\: \\ $$$${if}\:{n}=\mathrm{2}^{{k}} {a}\:\left({a}\:{is}\:{odd}\right)\:\left({k}\geqslant\mathrm{1}\right) \\ $$$${then}\:{q}=\mathrm{2}^{{k}} {b},\:{j}=\mathrm{2}^{{k}} {c} \\ $$$$\mathrm{2}^{{k}} {a}=\left(\mathrm{2}^{{k}} {c}\right)^{\mathrm{2}{kb}} \\ $$$$\mathrm{2}^{{k}} {a}=\left(\mathrm{2}^{{k}} \right)^{\mathrm{2}{kb}} {c}^{\mathrm{2}{kb}} \\ $$$${a}=\left(\mathrm{2}^{{k}} \right)^{\mathrm{2}{kb}−\mathrm{1}} {c}^{\mathrm{2}{kb}} \\ $$$${impossible}\:{since}\:{a}\:{is}\:{odd}\:{number} \\ $$$${so}\:{no}\:{even}\:{solutions}\:{for}\:{n}\:{for}\:{case}\:\left({i}\right) \\ $$$${continue}\:{for}\:{cases}\:{where}\:{n}\:{is}\:{odd} \\ $$

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

So no solution for n is possible for which if n is odd j must be odd and q even since n=j^q also n^2 =j^2 +q^2 j^(2q) =j^2 +q^2 j^2 (j^(2q−2) −1)=q^2 ⇒ ((q/j))^2 =j^(2q−2) −1 since j is odd and q even the above relation is impossible. correction: the above argument that q evrn and j odd is not correct. So the above relation may be possible. So no solution for n is possible for which satisfied both (i) and (ii).

$$\mathrm{So}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{n}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{for}\:\mathrm{which} \\ $$$${if}\:{n}\:{is}\:{odd}\:{j}\:{must}\:{be}\:{odd}\:{and}\:{q}\:{even} \\ $$$${since}\:{n}={j}^{{q}} \\ $$$${also} \\ $$$${n}^{\mathrm{2}} ={j}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$${j}^{\mathrm{2}{q}} ={j}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$${j}^{\mathrm{2}} \left({j}^{\mathrm{2}{q}−\mathrm{2}} −\mathrm{1}\right)={q}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\frac{{q}}{{j}}\right)^{\mathrm{2}} ={j}^{\mathrm{2}{q}−\mathrm{2}} −\mathrm{1} \\ $$$${since}\:{j}\:{is}\:{odd}\:{and}\:{q}\:{even}\:{the}\:{above}\:{relation} \\ $$$${is}\:{impossible}. \\ $$$${correction}:\:\mathrm{the}\:\mathrm{above}\:\mathrm{argument}\:\mathrm{that}\:{q}\:{evrn} \\ $$$${and}\:{j}\:{odd}\:{is}\:{not}\:{correct}.\:\:\mathrm{So}\:\mathrm{the}\:\mathrm{above} \\ $$$$\mathrm{relation}\:\mathrm{may}\:\mathrm{be}\:\mathrm{possible}. \\ $$$$\mathrm{So}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{n}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{for}\:\mathrm{which} \\ $$$$\mathrm{satisfied}\:\mathrm{both}\:\left({i}\right)\:{and}\:\left({ii}\right). \\ $$

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

for condition (iii) do you mean r,n,j,q can take −ve values as well.

$${for}\:{condition}\:\left({iii}\right)\:{do}\:{you}\:{mean}\:{r},{n},{j},{q}\:{can} \\ $$$${take}\:−{ve}\:{values}\:{as}\:{well}. \\ $$

Commented by Yozzii last updated on 19/Apr/16

Yes, but only can r be negative (or I originally posted for r to possibly be negative).

$${Yes},\:{but}\:{only}\:{can}\:{r}\:{be}\:{negative} \\ $$$$\left({or}\:{I}\:{originally}\:{posted}\:{for}\:{r}\:{to}\:{possibly}\right. \\ $$$$\left.{be}\:{negative}\right). \\ $$

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

j=2^b n=2^(bq) r=((n^2 −q^2 )/j^2 )=((2^(bq) −q^2 )/2^(2b) ) choose q=2^b ∙k r will be an integer ex. j=2^2 =4 q=2^2 ∙k say=2^2 ∙3=12 n=4^(12) r=((4^(24) −12^2 )/4^2 )=integer 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.

$${j}=\mathrm{2}^{{b}} \\ $$$${n}=\mathrm{2}^{{bq}} \\ $$$${r}=\frac{{n}^{\mathrm{2}} −{q}^{\mathrm{2}} }{{j}^{\mathrm{2}} }=\frac{\mathrm{2}^{{bq}} −{q}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}{b}} } \\ $$$$\mathrm{choose}\:{q}=\mathrm{2}^{{b}} \centerdot{k} \\ $$$${r}\:{will}\:{be}\:{an}\:{integer} \\ $$$${ex}. \\ $$$${j}=\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${q}=\mathrm{2}^{\mathrm{2}} \centerdot{k}\:{say}=\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}=\mathrm{12} \\ $$$${n}=\mathrm{4}^{\mathrm{12}} \\ $$$${r}=\frac{\mathrm{4}^{\mathrm{24}} −\mathrm{12}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }={integer} \\ $$$$\mathrm{So}\:\mathrm{infinite}\:\mathrm{solution}\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}. \\ $$$$\mathrm{Given}\:\mathrm{any}\:{j}\:\mathrm{we}\:\mathrm{can}\:\mathrm{always}\:\mathrm{choose}\:{q}\:\mathrm{such}\:\mathrm{that} \\ $$$${r}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}. \\ $$$${n}^{\mathrm{2}} >{q}^{\mathrm{2}} \:\mathrm{so}\:{r}\:\mathrm{will}\:\mathrm{always}\:\mathrm{be}\:+\mathrm{ve}. \\ $$

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