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Question Number 222516 by MrGaster last updated on 29/Jun/25

Given the integer k,how to find the incomplete general solution for the non-trivial integer solutions of the Diophantine equation: a^4 +b^4 +ka^2 b^2 =c^4 +d^4 +kc^2 d^2 ,a,b,c,d∈N,k∈Z,gcd(a,b,c,d)=1

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{integer}\:{k},\mathrm{how}\:\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{incomplete}\:\mathrm{general} \\ $$$$\mathrm{solution}\:\mathrm{for}\:\mathrm{the}\:\mathrm{non}-\mathrm{trivial}\:\mathrm{integer} \\ $$$$\mathrm{solutions}\:\mathrm{of}\:\:\mathrm{the}\:\mathrm{Diophantine}\:\mathrm{equation}: \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{ka}^{\mathrm{2}} {b}^{\mathrm{2}} ={c}^{\mathrm{4}} +{d}^{\mathrm{4}} +{kc}^{\mathrm{2}} {d}^{\mathrm{2}} ,{a},{b},{c},{d}\in\mathbb{N},{k}\in\mathbb{Z},\mathrm{gcd}\left({a},{b},{c},{d}\right)=\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 29/Jun/25

For k=2 (a^2 +b^2 )^2 =(c^2 +d^2 )^2 a^2 +b^2 =c^2 +d^2 ∨ a^2 +b^2 =−(c^2 +d^2 ) Rejected a=c ∧ b=d or a=d ∧ b=c Some of the general solutions for k=2: a=d,b=c or a=c,b=d For k=−2 (a^2 −b^2 )^2 =(c^2 −d^2 )^2 a^2 −b^2 =c^2 −d^2 ∨ a^2 −b^2 =−c^2 +d^2 a^2 +d^2 =b^2 +c^2 ∨ a^2 +c^2 =b^2 +d^2 (a=b ∧ c=d)∨ (a=c ∧ b=d) ∨ (a=d ∧ b=c) ...

$${For}\:{k}=\mathrm{2} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} +{d}^{\mathrm{2}} \:\vee\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =−\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\:{Rejected} \\ $$$${a}={c}\:\wedge\:{b}={d}\:{or}\:{a}={d}\:\wedge\:{b}={c} \\ $$$${Some}\:{of}\:{the}\:{general}\:{solutions}\:{for} \\ $$$${k}=\mathrm{2}: \\ $$$${a}={d},{b}={c} \\ $$$${or}\:{a}={c},{b}={d} \\ $$$${For}\:{k}=−\mathrm{2} \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={c}^{\mathrm{2}} −{d}^{\mathrm{2}} \:\vee\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =−{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{d}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:\vee\:{a}^{\mathrm{2}} +{c}^{\mathrm{2}} ={b}^{\mathrm{2}} +{d}^{\mathrm{2}} \\ $$$$\left({a}={b}\:\wedge\:{c}={d}\right)\vee\:\left({a}={c}\:\wedge\:{b}={d}\right) \\ $$$$\vee\:\left({a}={d}\:\wedge\:{b}={c}\right) \\ $$$$... \\ $$

Commented by MrGaster last updated on 29/Jun/25

Since there are k=2 and k=-2, what about other integers (such as k=1, -1, -3, -4, 4, etc.)?

Answered by Rasheed.Sindhi last updated on 29/Jun/25

a^4 +b^4 +ka^2 b^2 =c^4 +d^4 +kc^2 d^2 ,a,b,c,d∈N,k∈Z,gcd(a,b,c,d)=1 a^4 −c^4 +b^4 −d^4 =k(c^2 d^2 −a^2 b^2 ) (a−c)(a+c)(a^2 +c^2 )+(b−d)(b+d)(b^2 +d^2 )=k(cd−ab)(cd+ab) Special case: a=c and b=d satisfy the diophntine equation for every k: a−c=0 , b−d=0 ,cd−ab=ab−ab=0 (0)(a+c)(a^2 +c^2 )+(0)(b+d)(b^2 +d^2 )=k(0)(cd+ab) 0=0 (a,b,c,d)=(a,b,a,b) gcd(a,b,a,b)=1⇒gcd(a,b)=1 For example a=4,b=9 (a,b,c,d)=(a,b,a,b)=(4,9,4,9)

$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{ka}^{\mathrm{2}} {b}^{\mathrm{2}} ={c}^{\mathrm{4}} +{d}^{\mathrm{4}} +{kc}^{\mathrm{2}} {d}^{\mathrm{2}} ,{a},{b},{c},{d}\in\mathbb{N},{k}\in\mathbb{Z},\mathrm{gcd}\left({a},{b},{c},{d}\right)=\mathrm{1} \\ $$$${a}^{\mathrm{4}} −{c}^{\mathrm{4}} +{b}^{\mathrm{4}} −{d}^{\mathrm{4}} ={k}\left({c}^{\mathrm{2}} {d}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right) \\ $$$$\left({a}−{c}\right)\left({a}+{c}\right)\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\left({b}−{d}\right)\left({b}+{d}\right)\left({b}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)={k}\left({cd}−{ab}\right)\left({cd}+{ab}\right) \\ $$$${Special}\:{case}:\:{a}={c}\:{and}\:{b}={d}\:{satisfy} \\ $$$$\:{the}\:{diophntine}\:{equation}\:{for}\:{every}\:{k}: \\ $$$${a}−{c}=\mathrm{0}\:,\:{b}−{d}=\mathrm{0}\:,{cd}−{ab}={ab}−{ab}=\mathrm{0} \\ $$$$\left(\mathrm{0}\right)\left({a}+{c}\right)\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\left(\mathrm{0}\right)\left({b}+{d}\right)\left({b}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)={k}\left(\mathrm{0}\right)\left({cd}+{ab}\right) \\ $$$$\mathrm{0}=\mathrm{0} \\ $$$$\:\left({a},{b},{c},{d}\right)=\left({a},{b},{a},{b}\right) \\ $$$${gcd}\left({a},{b},{a},{b}\right)=\mathrm{1}\Rightarrow{gcd}\left({a},{b}\right)=\mathrm{1} \\ $$$${For}\:{example}\:{a}=\mathrm{4},{b}=\mathrm{9} \\ $$$$\:\left({a},{b},{c},{d}\right)=\left({a},{b},{a},{b}\right)=\left(\mathrm{4},\mathrm{9},\mathrm{4},\mathrm{9}\right) \\ $$$$\: \\ $$

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