`if a+b=c+d and a^2+b^2=c^2+d^2`, then show by mathematical induction `a^n+b^n=c^n+d^n`

Explanation

`P(n):a^n+b^n=c^n+d^n`
Step For `n=1 and n=2`,
`P(1):a+b=c+d and P(2):a^2+b^2+c^2+d^2` which are true (from given conditions) .
Therefore , `P(1) and P(2)` are true.
Step II Assume `P(k-1) and P(k) ` to the true
`therefore a^(k-1)+b^(k-1)=c^(k-1)+d^(k-1)` ..........(i)
and ` a^(k) +b^(k) =c^(k) +d^(k) `.......(ii)
Step III For `n=k+1`,
`P(k+1):a^(k+1)+b^(k+1)=c^(k+1)d^(k+1)`
`therefore LHS =a^(k+1)+b^(k+1)`
`=(a+b)(a^k+b^k)=ab^k-ba^k`
`=(a+b)(a^k+b^k)-ab(a^(k-1)+b^(k-1))" " ["Given " a+b=c+d and a^2+b^2=c^2+d^2, "then " ab =cd]`
`=(c+d)(c^k+d^k)-cd(c^k-1+d^k-1)` [From Eqs. (i) and (ii) ]
`=c^(k+1)+d^(k+1)=RHS`
Therefore `P(k+1)` is true. Hence by the principle of mathematical induction P(n) is true for all `n in N`.