Trying to prove or disprove this statement:
For all real numbers A, B, if $AB$ is rational and $A+B$ is rational, then $A^2$+$B^2$ is rational
Ive hit a wall as Im assuming A and B must either be rational or irrational in order for A+B to be rational (ex. A = $\sqrt2$ and B = $-\sqrt2$, or just A = 2 and B = 3), but that's not really getting me any further into proving or disproving the statement
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$\begingroup$Hint: $(A+B)^2=A^2+2AB+B^2,$ known as binomial theorem.
$\endgroup$ 2 $\begingroup$If $A+B = r\in \mathbb Q$ and $AB = q\in \mathbb Q$.
Then $A^2 + B^2 = (A+B)^2 - 2AB = r^2 -2q$.
Can we assume that if $r, q$ are rational that $r^2 -2q$ must be rational as well?
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If not.
Lemma 1: if $m,n\in \mathbb Q$ then $mn \in Q$.
Pf: Let $m = \frac ab; n =\frac cd, a,b,c,d \in \mathbb Z; b\ne 0, d\ne 0$. Then $mn = \frac {ac}{bd}$ and $ac,bd \in \mathbb Z$ and $bd \ne 0$. So $mn \in \mathbb Q$.
Lemma 2: if $m,n \in \mathbb Q$ then $m + n \in \mathbb Q$.
Pf: Let $m=\frac ab,n =\frac cd, a,b,c,d\in \mathbb Z;$ and $b,d\ne 0$. Then $m +n = \frac {ad + bc}{bd}$ and $ad+bc,bd\in \mathbb Z$ and $bd \ne 0$.
So $r^2 =r\times r \in \mathbb Q$ and $(-2)q \in \mathbb Q$ by Lemma 1. And $r^2 +(-2q)=r^2-2q\in \mathbb Q$ by Lemma 2.
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