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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