Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$ 
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$


*

*I managed to get $ab=cd$. Don't know how to proceed further.


Here's another method.
Note that by re-arranging, we may write $a+b=c+d$ either as $a-c=d-b$ or as $b-c = d-a.$  Just keep that in mind as we manipulate the second equation:
$$\begin{align*}0 &= (c^2 + d^2) - (a^2 + b^2) \\ &= (d^2 - b^2) - (a^2 - c^2) \\ &= (d-b)(d+b) - (a-c)(a+c) \\ &= (a-c)(d+b) - (a-c)(a+c) \\ &= (a-c)((d+b)-(a+c)) \\ &= (a-c)((d-a)+(b-c)) \\ &= (a-c)((d-a)+(d-a)) \\ &= 2(a-c)(d-a)\end{align*}$$
[Do you see the steps I've used each of $a-c=d-b$ and $b-c=d-a$ to simplify the expression?]

At this point, we're effectively done, as the factorization gives us only two solutions:

*

*$a-c=0$: Then $d-b=a-c = 0$ as well, which gives $a=c$ and $b=d$.

*$d-a=0$: Then $b-c=d-a = 0$ as well, which gives $a=d$ and $b=c$.

