From a+b=c+d

=> (a+b)²=(c+d)²

=>ab=cd

Consider (a+b)³=(c+d)³

We have a³+3ab(a+b)+b³=c³+3cd+d³

=>a³+b³=c³+d³

Consider (a³+b³).(a+b)=(c³+d³).(c+d)

=> a⁴+ab(a²+b²)+b⁴=c⁴+cd(c²+d²)

=>a⁴+b⁴=c⁴+d⁴

We can infer this

aⁿ+bⁿ=cⁿ+dⁿ

=> (aⁿ+bⁿ).(a+b)=(cⁿ+dⁿ).(c+d)

=> aⁿ⁺¹+ab(a^n-1+b^n-1)+bⁿ⁺¹=cⁿ⁺¹+cd(c^n-1+d^n-1)+dⁿ⁺¹

=>aⁿ⁺¹+bⁿ⁺¹=cⁿ⁺¹+dⁿ⁺¹

So we have a²⁰¹⁸+b²⁰¹⁸=c²⁰¹⁸+d²⁰¹⁸

From a+b=c+d

=> (a+b)2=(c+d)2

=>ab=cd

Consider (a+b)3=(c+d)3

We have a3+3ab(a+b)+b3=c3+3cd+d3

=>a3+b3=c3+d3

Consider (a3+b3).(a+b)=(c3+d3).(c+d)

=> a4+ab(a2+b2)+b4=c4+cd(c2+d2)

=>a4+b4=c4+d4

We can infer this

an+bn=cn+dn

=> (an+bn).(a+b)=(cn+dn).(c+d)

=> an+1+ab(an-1+bn-1)+bn+1=cn+1+cd(cn-1+dn-1)+dn+1

=>an+1+bn+1=cn+1+dn+1

So we have a2018+b2018=c2018+d2018

From a+b=c+d

=> (a+b)2=(c+d)2

=>ab=cd

Consider (a+b)3=(c+d)3

We have a3+3ab(a+b)+b3=c3+3cd+d3

=>a3+b3=c3+d3

Consider (a3+b3).(a+b)=(c3+d3).(c+d)

=> a4+ab(a2+b2)+b4=c4+cd(c2+d2)

=>a4+b4=c4+d4

We can infer this

an+bn=cn+dn

=> (an+bn).(a+b)=(cn+dn).(c+d)

=> an+1+ab(an-1+bn-1)+bn+1=cn+1+cd(cn-1+dn-1)+dn+1

=>an+1+bn+1=cn+1+dn+1

So we have a2018+b2018=c2018+d2018