E : 17 056 Everything is just zero

           It is Problem E:17056 of GMB 11/2024, page 651, authors Dragoș PETRICĂ and Cosmin MANEA from Pitești. In translation:

              "Determine the natural numbers $x,\;y,\;z,\;t\;$ and $u\;$ that simultaneously 

                 verify the relations  $x^2+y^2+z^2+t^2=7\cdot u^2$  and  $x\cdot t=y\cdot z$"

 ANSWER CiP

$$x=y=z=t=u=0$$

                       Solution CiP

               Taking into account the relationship

$$x\cdot t=y\cdot z \tag{1}$$

 the given equation is written successively equivalent to

$(x^2\pm 2 \cdot xt+t^2)+(y^2\mp 2 \cdot yz+z^2)=7\cdot u^2\;\;\Leftrightarrow$

$$\Leftrightarrow\;\;(x\pm t)^2+(y\mp z)^2=7\cdot u^2. \tag{2}$$

From (2) it is seen that the prime number $7$ divides a sum of two squares of integers. We will first prove the following

           Lemma  For  $A,\;B \in \mathbb{Z}$  we have   $7\;\mid\;A^2+B^2\;\;\Rightarrow\;\;7\mid A\;\;and\;\;7\mid B.$

            Proof of the Lemma 

            We have, after the remainders of dividing $A$ by $7$, one of the possibilities

$$(i)\;A=7 \cdot k,\;\;(ii)\;A=7\cdot k\pm 1,\;\;(iii)\;A=7 \cdot k \pm 2,\;\;(iv)\;A=7\cdot k \pm3,\;\;\; k\in \mathbb{Z}$$

Then, we have respectively $(i)\;A^2=(7k)^2=7\cdot(7k^2)=7\cdot K$

$(ii)\;A^2=(7k\pm 1)^2=49k^2\pm 14k+1=7\cdot K+1$

$(iii)\;A^2=(7k\pm 2)^2=49k^2\pm 28k+4=7\cdot K+4=7\cdot(K+1)-3$

$(iv)\;A^2=(7k\pm 3)^2=49k^2 \pm 42k+9=7\cdot (7k^2\pm 6k+1)+2$

where, in different cases, the number $K$ has different values. In other words, the natural remainders of dividing $A^2$ by $7$ can only be  $0,\;1,\;2\; or\; 4.$ Considering instead of the remainder $4$ the equivalent value $-3$, $(mod\; 7)$, we have one of the possibilities

$$A^2=7\cdot K,\;\;A^2=7 \cdot K+1,\;\;A^2=7\cdot K+2,\;\;A^2=7\cdot K-3,\;\;\;K\in\mathbb{N}.$$

     Obviously we have the same cases for $B^2$, and then, for $A^2+B^2$  we have the possibilities in the Table below, also calculated mod_7:

$$\begin{array}{|c|c|c|c|c|}\hline B^2/A^2&0&1&2&-3\\ \hline  0&0&1&2&-3\\ \hline  1&1&2&3&-2\\ \hline 2&2&3&-3&-1\\ \hline -3&-3&-2&-1&2\\ \hline\end{array}$$

            It can be seen from the Table above that $7\mid A^2+B^2$, that is $A^2+B^2=0\;mod\;7$ if and only if $A^2=0\;mod\;7\;and\;\;B^2=0\;mod\;7$, or $7\mid A^2\;\;and\;\;7\mid B^2$, and hence $7\mid A\;\;and\;\;7\mid B.$

$\square\;Lemma$

 

          Returning to (2), we have, according to Lemma

$$|x\pm t|=7\cdot v,\;\;|y\mp z|=7\cdot w,\;\;\;v,\;w \in \mathbb{N}\;;\;\Rightarrow$$

$\Rightarrow\;49(v^2+w^2)=7u^2\;\;\Leftrightarrow\;\;7(v^2+w^2)=u^2$, and hence $7\mid u$, so $u=7\dot u_1$ and

$$v^2+w^2=7\cdot u_1^2,\;\;\;u_1<u,\;\;u_1\in\mathbb{N}.\tag{3}$$

But in equation (3) we see that $7\mid (v^2+w^2)$ so, according Lemma again, $7\mid v$ and $7\mid w$. Then $v=7\cdot v_1,\;w=7\cdot w_1$ and $$49 \cdot v_1^2+49 \cdot w_1^2=7\cdot u_1^2\;\;\Leftrightarrow\;\;7(v_1^2+w_1^2)=u_1^2.$$

From here it follows $7\mid u_1$ so $u_1=7\cdot u_2$, and

$$v_1^2+w_1^2=7 \cdot u_2^2,\;\;\;u_2<u_1,\;\;u_2\in\mathbb{N}. \tag{4}$$

Comparing (3) and (4) we see that, continuing the process, we would obtain an infinitely decreasing sequence of natural numbers $u>u_1>u_2>\dots$, which is not possible. So $u=0.$

     But then $x^2+y^2+z^2+t^2=0$ and hence $x=y=z=t=0$. We got the answer.

 $\blacksquare$

          Remark CiP  Although $7$ is a sum of four squares $7=2^2+1^1+1^2+1^2$  the condition (1) is not met.

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