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GAZETA MATEMATICĂ Seria B N0 11/2022

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SUPLIMENTUL cu EXERCIȚII al GMB N0 11/2022

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Happy_? New Year 2023 ??

           The question posed by Marcel Țena on Facebook sparked heated debates:

          "How do you write 2023 as the sum of four perfect squares ?"

Without finding an answer, I commented with a challenge. "Petre Ciobanu Challenge: I learned, I don't know where, that two of the four numbers can always be chosen equally. Who? find this writing ??"

However, the comment did not have any reaction, and I also replied, later: 

So $2023=17^2+23^2+23^2+26^2$. I also mentioned Preda Mihailescu in the comment, who I think is the one who demonstrated the property that in Lagrange's Four Squares Theorem, two numbers can always be chosen equal.

     Mr. Vasile MASGRAS published the following list

in total 42 representations.

     Mr. Iulian MUSCALU sent to a link with 109 representations.

 

Among the 109 representations, some are identical. I counted 64, finding some identical ones (marked with black). Then, checking with the correct list of the 61 representations (put by Eugen Paltanea see also link), I found three more that are repeated (marked in green). The leftmost number (written in green) is the number corresponding to the correct table with the 61 representations.

     The correct answer is therefore 61 representations of the number 2023 as the sum of four perfect squares.

 Of them, 42 variants contain all four distinct numbers. Then 18 variants contain two equal numbers and one has a number repeated three times (marked with red color)

          In agreement with Jacobi's Theorem, the number of ways to represent n as the sum of four squares is eight times the sum of the divisors of n if n is odd. Here

$$8*(1+7+17+119+289+2023)=8*2456=19\;648.$$

      Let's count them.

     A writing with 4 different terms, e.g. $2023=1^2+2^2+13^2+43^2$, is counted, taking into account all $\pm$ possible signs

 $$2023=(\pm 1)^2+(\pm 2)^2+(\pm 13)^2+(\pm 43)^2,$$

 $2*2*2*2=16$ times, each writing is counted 24 times (the number of arranging the 4 terms, e.g. 2023=43^2+1^2+2^2+13^2). In total $24*16=384$. Most are of this type, $42*384=16\;128.$

     Writings with two equal terms, e.g. $2023=1^2+10^2+31^2+31^2$, count 16 times if we take into account the $\pm$ signs, and 12 times each if we take into account all possible arrangements. There are 18 of these in total. So we have in total $18*12*16=3\;456$ writings.

      Finally, the only writing with three equal terms, $2023=17^2+17^2+17^2+34^2$ is counted 16 times because of the signs $\pm$, and 4 times because of the permutations of the terms. In total $16*4=64.$

        In conclusion, the total number of writings will be $$16128+3456+64=19648.$$

Problem 28423 from GMB 10/2022

            The problem is proposed for the 9th grade, on page 483.

In translation, thanks to Miss Google:

          "Show that the equation $x^{n+1}+y^n=z^n$ has an infinity of solutions in the set of nonzero natural numbers, for any natural number $n \geqslant 2$."

ANSWER CiP

$$x=\frac{(k^{n+1}+1)^n-1}{k^n},\;\;y=\frac{(k^{n+1}+1)^n-1}{k^{n+1}},\;\;z=\frac{(k^{n+1}+1)^{n+1}-k^{n+1}-1}{k^{n+1}}$$

represent solutions of the given equation, for $k \in \mathbb{N}\setminus \{0\}.$

                    Solution CiP

               Let us first show that the values given in the Answer are solutions of the equation.

          Let $l \underset{def}{=} k^{n+1}+1.$ Then $x=\frac{l^n-1}{k^n},\;y=\frac{l^n-1}{k^{n+1}},\;z=\frac{l^{n+1}-l}{k^{n+1}}$ and we have

$$x^{n+1}+y^n=\frac{(l^n-1)^{n+1}}{k^{n(n+1)}}+\frac{(l^n-1)^n}{k^{(n+1)n}}=$$ 

$=\frac{(l^n-1)^n}{k^{(n+1)n}}\cdot [(l^n-1)+1]=\frac{(l^n-1)^n \cdot l^n}{k^{(n+1)n}}=\left ( \frac{l(l^n-1)}{k^{n+1}} \right )^n=\left ( \frac{l^{n+1}-l}{k^{n+1}} \right ) ^n\;=\;\;z^n.$

          We found the expressions in the answer by trying to find solutions of the given equation with $x=ky,\;k \in \mathbb{N} \setminus \{ 0 \}.$ We have

 $$x^{n+1}+y^n=k^{n+1} \cdot y^{n+1}+y^n=y^n(k^{n+1} \cdot y+1).$$ 

For the above result to be a power of $n$ we must have $k^{n+1}y+1=l^n$ for some integer $l$, so $y=\frac{l^n-1}{k^{n+1}}.$ For $y$ to be an integer, it is enough to choose $l=k^{n+1}+1$ because according to the binomial theorem we have

$l^n-1=(k^{n+1}+1)^n-1=\sum_{i=0}^n \textrm{C}_n^i (k^{n+1})^{n-i}=$ 

$=[(k^{n+1})^n+\textrm{C}_n^1(k^{n+1})^{n-1}+ \cdots +\textrm{C}_n^{n-1}k^{n+1}+1]-1=k^{n+1} \cdot A$,

 for some integer $A$. That's how I found the answer.

$\blacksquare$

           REMARK CiP

          In particular cases, the solutions can be expressed differently. For $n=2$ the equation

$$x^3+y^2=z^2$$

$\Leftrightarrow \;x^3=(z-y)(z+y)$ can be solved by choosin

$$\begin{cases}z-y=x\\z+y=x^2\end{cases}$$

obtaining an infinite family of complete solutions $$x=m,\;y=\frac{m^2-m}{2},\;z=\frac{m^2+m}{2},\;m\in \mathbb{Z}.$$

(I did not investigate if these represent the complete solutions of the equation.)

          The particular case has the following solutions

The general formula provides the solutions in the case of n=2

Problem E:16386 from GMB 10/2022

          The problem proposed for class 6, on page 480.

In translation (thanks to Miss Google):

          "Let $p$ and $q$ be two distinct prime numbers. Prove that if  $15(p+q) >8pq$, then 

          $$93pq \leqslant 174(p+q) \leqslant 145  pq."$$

 

ANSWER CiP

           One of the numbers is $2$

 and the other can be $3,\;5,\;7,\;11,\;13,\;17,\;19,\;23,\;29.$

             SOLUTION CiP

          Natural numbers that verify the condition

$$15(p+q)>8pq \tag{1}$$

 must be in finite quantity. Indeed, the condition (1) can be written equivalent to

$$\frac{1}{p}+\frac{1}{q}>\frac{8}{15}. \tag{2}$$

     If none of the numbers $p,q$, is equal to $2$, then $\frac{1}{p}+\frac{1}{q} \leqslant \frac{1}{3}+\frac{1}{5}=\frac{8}{15},$ contradicting (2). 

      Assuming $p=2$ the condition (1) is written $15(2+q)>16q\;\;\Leftrightarrow\;\;q<30$. So $3\leqslant q \leqslant 29$ and $q$ is a prime number. Then we have 

$$\frac{1}{2}+\frac{1}{29}\leqslant \frac{1}{p}+\frac{1}{q}\leqslant \frac{1}{2}+\frac{1}{3}\;\Leftrightarrow\;\frac{31}{58}\leqslant \frac{p+q}{pq}\leqslant \frac{5}{6}$$

and multiplying the last inequalities by $6 \cdot 29=174$ we get the conclusion.

$\blacksquare$

          REMARK CiP

          The condition (1) can also be written, after multiplying by 8, $$64pq-120p-120q<0\;\Leftrightarrow$$

$\Leftrightarrow \;(8p-15)(8q-15)<225$. If $p\geqslant 3$, we have $8p-15 \geqslant 9$ then we deduce $8q-15<\frac{225}{8p-15}\leqslant \frac{225}{9}=25$, so $8q<40$, $q<5$. We only have the possibility $q=3$ or $q=2$. When $q=3$, from (1) we deduce $p<5$. We can only have $p=3$ but then $p=q$, impossible. When $q=2$, from (1) we deduce $p<30$ and we find the answer.

$\square$ end REM

GAZETA MATEMATICĂ Seria B N0 10/2022

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