Problem E:15743 GMB 6-7-8/2020, pag 367

 Proposed for the seventh grade. In translation:

          "Let $a$ and $b$ be real numbers such that $a+b$, $a^{3}+b^{3}$ and $a+b^{2}$ are nonzero rational numbers.

     a) Give an example of irrational numbers $a$ and $b$ that check the relationships in the statement.

     b) Prove that $a^{2}+b$ is a rational number."

 

ANSWER CiP

a) $a=\frac{1+\sqrt{2}}{2}$,   $b=\frac{1-\sqrt{2}}{2}$;

          a family of numbers with the indicated properties is

 $a=\frac{1}{2}+\frac {1}{2}\cdot \sqrt{d}$, $b=\frac{1}{2}-\frac{1}{2} \cdot \sqrt{d}$,

          with $d$ rational number that is not a perfect square.

 

Solution CiP

           a) We calculate that the values indicated in the answer check the conditions of

 the problem:

$a+b=(\frac{1}{2}+\frac{1}{2}\cdot \sqrt{d})+(\frac{1}{2}-\frac{1}{2}\cdot \sqrt{d})=1$,

$a^{3}+b^{3}=\left ( \frac{1}{2}+\frac{1}{2} \cdot \sqrt{d} \right )^{3}+\left (\frac{1}{2}-\frac{1}{2} \cdot \sqrt{d} \right )^{3}=$

$=(\frac{1}{8}+\frac{3}{8}\sqrt{d}+\frac{3}{8}d +\frac{1}{8}d\sqrt{d})+(\frac{1}{8}-\frac{3}{8}\sqrt{d}+\frac{3}{8}d-\frac{1}{8}d \sqrt{d})=\frac{1}{4}+\frac{3}{4}d$,

$a+b^{2}=(\frac{1}{2}+\frac{1}{2}\cdot \sqrt{d})+(\frac{1}{2}-\frac{1}{2}\cdot \sqrt{d})^{2}=$

$=\frac{1}{2}+\frac{1}{2} \cdot \sqrt{d}+(\frac{1}{4}-\frac{1}{2} \cdot \sqrt{d}+\frac{1}{4}\cdot d)=\frac{3}{4}+\frac{1}{4}d$.

          b) Let's write down three numbers $r,s,p$ with the property

\begin{cases}a+b=r\;\;\;\;\;(1)\\a^{3}+b^{3}=p\;\;(2)\;\;\;\;\;\;\;r,p,s \in \mathbb{Q}\setminus \{0 \}\\a+b^{2}=s\;\;\;\;(3) \end{cases}

           Note that a necessary condition for the existence of numbers $a,b$ is

$(4)$                    $r-s \leqslant \frac{1}{4}$.

 Because, assuming that $a,b$ exist, we have

$-\left ( b-\frac{1}{2} \right )^{2}\leqslant 0 \;\Leftrightarrow \; b-b^{2} \leqslant \frac{1}{4}\;\Leftrightarrow \; (a+b)-(a+b^{2}) \leqslant \frac{1}{4}\;\overset{(1),(3)}\Leftrightarrow\;r-s\leqslant \frac{1}{4}.$

         

                A "negligent" demonstration

     
          Using the equation $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$, it results from (2) and (1),    $a^{2}-ab+b^{2}=\frac{p}{r}$ $\Rightarrow$ $(a+b)^{2}-3ab=\frac{p}{r}$ $\Rightarrow$ $ab=\frac{r^{2}}{3}-\frac{p}{3r} \in \mathbb{Q}$.

Further $\Rightarrow a^{2}+b^{2}=\frac{p}{r}+ab=\frac{p}{r}+\frac{r^{2}}{3}-\frac{p}{3r}=\frac{r^{2}}{3}+\frac{2p}{3r} \in \mathbb{Q}$.

     From (3), $s=a+b^{2} \overset{(1)}=(r-a)+(\frac{r^{2}}{3}+\frac{2p}{3r}-a^{2})=\frac{r^{2}}{3}+r+\frac{2p}{3r}-(a^{2}+b)$  

$\Rightarrow$  $a^{2}+b=\frac{r^{2}}{3}+r+\frac{2p}{3r}-s \in \mathbb{Q}$.

$\blacksquare$

 

                Remarks at the demonstration

           $\blacklozenge$ The numbers $r,\;p,\;s$ that fulfill the relationships (1)-(3) can be independent ...

          $\blacklozenge$ Eliminating $a$ from the equations (1) and (3) leads to the equation $b^{2}-b=s-r$. From here we get two values for $b$, and finally

$(5)$          $a=\frac{2r-1\pm \sqrt{1-4r+4s}}{2}$    $b=\frac{1 \mp \sqrt{1-4r+4s}}{2}$.

(From $\pm, \mp$ the upper signs are taken once, then the lower ones.)

 

           $\blacklozenge$ A relationship between $r,\;p,\;s$ is obtained by replacing formulas (5) in formula (2)

$p=a^{3}+b^{3}=(r-\frac{1}{2}\pm \frac{\sqrt{1-4r+4s}}{2})^{3}+(\frac{1}{2}\mp \frac{\sqrt{1-4r+4s}}{2})^{3}$.

Without completing the calculations, we find at some point

$(6)$   $p=(r-\frac{1}{2})^{3}\pm \frac{3}{2}(r^{2}-r)\sqrt{1-4r+4s}+\frac{1}{8}+\frac{3}{4}(1-4r+4s)$.

     If $a$ and $b$ are irrational numbers, then $1-4r+4s$ is not the square of a rational number (see (5)). Then (6) shows that, in order to have $p \in \mathbb{Q}$ it must $r^{2}-r=0$, so $r=1$ (the case $r=0$ being excluded from the statement).

          $\blacklozenge$ We see now from (5) that condition (4) is also sufficient for the equations (1)-(3) to admit real solutions. However, this only happens if the value of $p$ is given by (6); the numbers $r$ and $s$ can be given arbitrarily (of course respecting the relation (4)).

 

 

          $\blacklozenge$  We return to case $r=1$, the only one in which the equations (1)-(3) admit irrational numbers as solutions.

 

\begin{cases}a+b=1\;\;\;\;(7)\\a^{3}+b^{3}=p\;(8)\\a+b^{2}=s\;\;\;(9) \end{cases}

 We have in (9) condition $s \geqslant \frac{3}{4}$ (deduced from (4)). The values $a$ and $b$ given now by (5) are

$\left \{ a,\;b \right \}=\left \{ \frac{1\pm \sqrt{4s-3}}{2} \right \}$,

 and the calculation that led to relation (6) provides us $p=3s-2$. (Because $s \geqslant \frac {3}{4}$ we have $p \geqslant \frac{1}{4}$.) 

     Therefor, we have the (!equivalent) possibilities

\begin{cases} a+b=1\\a^{3}+b^{3}=3s-1\;\;\;s\geqslant \frac{3}{4}\\a+b^{2}=s \end{cases}

or 

\begin{cases}a+b=1\\a^{3}+b^{3}=p\;\;\; p\geqslant \frac{1}{4}\\a+b^{2}=\frac{p+2}{3} \end{cases}

with solutions

 $\left \{a,\;b \right \}=\left \{\frac{1\pm \sqrt{4s-3}}{2}\right \}=\left \{\frac{1\pm \sqrt{\frac{4p-1}{3}}}{2} \right \}$.

 

           $\blacklozenge$ $p$ depending on $r$ and $s$

          It is the relation (6), completed

$(6^{bis})$     $p=r^{3}-\frac{3}{2}r^{2}-\frac{3}{4}r+1+3s \pm \frac{3}{2}(r^{2}-r)\sqrt{1-4r+4s}\;$.

 

          $\blacklozenge$ $s$ depending on $p$ and $r$

           We will show below that we have

$(10)$          $s=\frac{1}{6}r^{2}+\frac{1}{2}r+\frac{p}{3r}\pm \frac{1}{2}(r-1)\sqrt{\frac{4p-r^{3}}{3r}}$.

      For this, consider only equations (1) and (2).From these we have obtained befor

$a+b=r$     $a\cdot b=\frac{r^{3}-p}{3r}$.

 So $a$ and $b$ will be the roots of a quadratic equation $x^{2}-(a+b)\cdot x +ab=0$. We find

$\left \{ a,b \right \}=\left \{ r\pm \sqrt{\frac{4p-r^{3}}{2}} \right \}$.

With these values we calculate $a+b^{2}$ and obtain the relation (10).

          $\blacklozenge$ $r$ depending on $p$ and $s$

      $(11)$     $r^{6}-3r^{4}(s-1)-r^{3}(2p+9s)+9r^{2}(s^{2}+p)-3rp(2s+1)+p^{2}=0$.

          This is obtained by writing the given relationships (1)-(3) in the form \begin{cases}b=r-a\\b^{2}=s-a\\b^{3}=p-a^{3}.\end{cases}

 From the last two relations we obtain successively 

$\begin{cases} (r-a)^{2}=s-a\\(r-a)^{3}=p-a^{3} \end{cases}$ $\Leftrightarrow$ $\begin{cases}a^{2}-(2s-1)\cdot a +r^{2}-s=0\\3r\cdot a^{2}-3r^{2} \cdot a +r^{3}-p=0. \end{cases}$

It is known that two quadratic equations

$\alpha \cdot x^{2}+\beta \cdot x+\gamma =0$ and $\alpha_{1} \cdot x^{2}+\beta_{1} \cdot x +\gamma_{1}=0$

 have a common root if, and only if

$(\alpha \cdot \gamma_{1}-\gamma \cdot \alpha_{1})^{2}=(\alpha \cdot \beta_{1}-\beta \cdot \alpha_{1})\cdot (\beta \cdot \gamma_{1}-\gamma \cdot \beta_{1})$.

Here it is written

$(12)$                    $(2r^{3}-3rs+p)^{2}=3r(r-1)(r^{4}+r^{3}-3r^{2}s+2rp-p)$

 from which it results (11).

          $\blacklozenge$   We can also arrange the formula (12) in the forms

$9r^{2} \cdot s^{2}-3r(r^{3}+3r^{2}+2p) \cdot s+r^{6}+3r^{4}-2r^{3}p+9r^{2}p-3rp+p^{2}=0$,

 where we find equation (10) again,

or,

$p^{2}-p\cdot r(2r^{2}-9r+6s+3)+r^{2}[r^{4}-3r^{2}(s-1)-9rs+9s^{2}]=0$,

where we find again $(6^{bis})$.

$\blacklozenge$  $\blacklozenge$  $\blacklozenge$