a) Determine the parameters $p, q$ such that the expressions

$$
(x+3)^{2}+(7 x+p)^{2} \text { and }(3 x+5)^{2}+(p x+q)^{2}
$$

are equal to the square of a first-degree polynomial.

b) Let $a, b, c, A, B, C$ be given, non-zero numbers such that the polynomials

$$
(a x+b)^{2}+(A x+B)^{2} \text { and }(b x+c)^{2}+(B x+C)^{2}
$$

are equal to the square of a first-degree polynomial. Show that the same holds for the following polynomial:

$$
(c x+a)^{2}+(C x+A)^{2}
$$
I. solution. It is generally useful to determine under what conditions the sum of the squares of two linear expressions is equal to the square of a linear expression, because we can use this in the answers to each question. Let $\alpha, \beta, \gamma, \delta$ be any four numbers, and consider the polynomial

$$
(\alpha x+\beta)^{2}+(\gamma x+\delta)^{2}=\left(\alpha^{2}+\gamma^{2}\right) x^{2}+2(\alpha \beta+\gamma \delta) x+\beta^{2}+\delta^{2}
$$

If here $\alpha=\gamma=0$, then we are dealing with a non-negative constant, which is indeed the square of a constant. If this is not the case, then complete the square for the expression:

$$
\begin{gathered}
\left(\sqrt{\alpha^{2}+\gamma^{2}} x+\frac{\alpha \beta+\gamma \delta}{\sqrt{\alpha^{2}+\gamma^{2}}}\right)^{2}+\beta^{2}+\delta^{2}-\frac{(\alpha \beta+\gamma \delta)^{2}}{\alpha^{2}+\gamma^{2}}=\left(\sqrt{\alpha^{2}+\gamma^{2}} x+\frac{\alpha \beta+\gamma \delta}{\sqrt{\alpha^{2}+\gamma^{2}}}\right)^{2}+ \\
\frac{\alpha^{2} \delta^{2}+\beta^{2} \gamma^{2}-2 \alpha \beta \gamma \delta}{\alpha^{2}+\gamma^{2}}
\end{gathered}
$$

The expression is the square of a linear expression if and only if the last fraction is 0. The numerator of this fraction is $(\alpha \delta-\beta \gamma)^{2}$, so our quadratic polynomial is the square of a linear expression if and only if

$$
\alpha \delta=\beta \gamma
$$

(This is also true in the case $\alpha=\gamma=0$, so more precisely, we should say "at most quadratic" and "at most linear" polynomials.)
a) For the first expression under (1),

$$
\alpha=1, \quad \beta=3, \quad \gamma=7, \quad \delta=p
$$

thus (4) gives the value $p=21$. For the second expression,

$$
\alpha=3, \quad \beta=5, \quad \gamma=p, \quad \delta=q
$$

so from (4), $3 q=5 p=105$, i.e., $q=35$. Indeed,

$$
\begin{aligned}
& (x+3)^{2}+(7 x+21)^{2}=(\sqrt{50} x+3 \sqrt{50})^{2} \quad \text { and } \\
& (3 x+5)^{2}+(21 x+35)^{2}=(3 \sqrt{50} x+5 \sqrt{50})^{2}
\end{aligned}
$$

b) The two expressions under (2) are, according to (4), the square of at most a linear expression if and only if

$$
a B=b A \quad \text { and } \quad b C=c B
$$

Multiplying these two conditions,

$$
a b B C=b c A B, \quad \text { i.e., } \quad b B(a C-c A)=0
$$

Since by condition $b$ and $B$ are not 0, then

$$
a C-c A=0, \quad a C=c A
$$

and from this, according to (4), it follows that the expression (3) is also the square of at most a linear expression.

András Csikvári (Budapest, Berzsenyi D. Gymnasium, 2nd year)

II. solution. Examine the function generated by the individual expressions for $x$, and how they can become 0. The sum of two squares can only be 0 if both squares are 0, and the square of a linear expression disappears for the value of $x$ that makes the base 0. The problem thus wants to find, in each case, the value of $x$ for which the bases of the two added squares are zero.

a) For the first and second expressions under (1), the corresponding $x$ values are

$$
x_{1}=-3=-\frac{p}{7} \quad \text { and } \quad x_{2}=-\frac{5}{3}=-\frac{q}{p}
$$

and from this, as in the first solution, the values

$$
p=21, \quad q=\frac{5}{3} p=35
$$

are obtained.

b) For the expressions under (2), these values are

$$
x_{3}=-\frac{b}{a}=-\frac{B}{A} \quad \text { and } \quad x_{4}=-\frac{c}{b}=-\frac{C}{B}
$$

If these relationships hold, then by multiplying them, we get

$$
\frac{c}{a}=\frac{C}{A}
$$

Denoting the common value of these two fractions by $\lambda$, and expressing $c$ and $C$ from this, the polynomial (3) becomes:

$$
(\lambda a x+a)^{2}+(\lambda A x+A)^{2}=\left[\sqrt{a^{2}+A^{2}}(\lambda x+1)\right]^{2}
$$

which is indeed the square of a linear expression.

Using the work of István Dettai (Pannonhalma, Bencés Gymnasium, 2nd year)