6.253. Form a quadratic equation with roots $(a+b)^{2}$ and $(a-b)^{2}$, if $a$ and $b$ are the roots of the equation $x^{2}+p x+q=0$.
## Solution.

Let's form the quadratic equation $y^{2}+P y+Q=0$.

By Vieta's theorem and the condition, we have

$$
\begin{aligned}
& \left\{\begin{array}{l}
P=-\left((a-b)^{2}+(a+b)^{2}\right)=-\left(a^{2}-2 a b+b^{2}+a^{2}+2 a b+b^{2}\right), \\
Q=(a-b)^{2} \cdot(a+b)^{2}
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
P=-2\left(a^{2}+b^{2}\right)=-2\left((a+b)^{2}-2 a b\right), \\
Q=(a-b)^{2} \cdot(a+b)^{2} .
\end{array}\right.
\end{aligned}
$$

To find the values of $a+b, a-b$, and $a b$, we use Vieta's theorem again:

$$
\left\{\begin{array} { l } 
{ a + b = - p , } \\
{ a b = q }
\end{array} \Rightarrow \left\{\begin{array}{l}
(a+b)^{2}=p^{2}, \\
a b=q
\end{array}, \Leftrightarrow\left\{\begin{array}{l}
a^{2}+2 a b+b^{2}=p^{2}, \\
a b=q
\end{array} \Leftrightarrow\right.\right.\right.
$$

$\Leftrightarrow\left\{\begin{array}{l}a^{2}+2 a b+b^{2}=p^{2}, \\ -4 a b=-4 q .\end{array} \Rightarrow a^{2}-2 a b+b^{2}=p^{2}-4 q,(a-b)^{2}=p^{2}-4 q\right.$.

Thus, $\left\{\begin{array}{l}P=-2\left(p^{2}-2 q\right), \\ Q=\left(p^{2}-4 q\right) p^{2} .\end{array}\right.$

The desired equation is $y^{2}-2\left(p^{2}-2 q\right) y+\left(p^{2}-4 q\right) p^{2}=0$.

Answer: $y^{2}-2\left(p^{2}-2 q\right) y+\left(p^{2}-4 q\right) p^{2}=0$.