momo1729 wrote: Let $a, b, c, d, m, n$ be positive integers such that $a^{2}+b^{2}+c^{2}+d^{2}=1989$ and $n^{2}=max\left \{ a,b,c,d \right \}$. Find the values of $m$ and $n$. Wlog say $\max(a,b,c,d)=a$ and so we have $n^4+b^2+c^2+d^2=1989<7^4$ And since $1989=6^4+26^2+4^2+1^2$ we get the answer $\boxed{n=6}$ And I have no idea at all about what could be the value of $m$ which is never defined in your problem statement.

momo missed the other important condition : a+b+c+d=m².

Thank you Dijkschneier. I'm editing the topic.

The fact that $a^2 + b^2 + c^2 + d^2 = 1989 \equiv 1 \pmod{4}$ implies that of the four integers $a$, $b$, $c$ and $d$, three are even and one is odd. Hence $m^2 = a+b+c+d$ is odd, implying $m$ is odd. Since $a$, $b$, $c$ and $d$ are symmetric in the two given equations, we may assume $a \leq b \leq c \leq d$. Hence $d = max \{a,b,c,d\} = n^2$. Now $n^4 = d^2 < a^2 + b^2 + c^2 + d^2 = 1989 \leq 4d^2 = 4n^4$, implying $n \in \{5,6\}$. Futhermore $n^2 = d < m^2 = a + b + c + d \leq 4d = (2n)^2$, i.e. $(1) \; n < m < 2n$ since $m$ is odd. Assume $n=5$. Then $(2) \; a^2 + b^2 + c^2 = 1989 - 5^4 = 1989 - 625 = 1364$. The fact that 1364 is divisible by 4 combined with (2) implies $a$, $b$ and $c$ are even, i.e. $a=2x$, $b=2y$ and $c=2z$ for three positive integers $x, y$ and $z$. Consequently $x^2 + y^2 + z^2 = 341 \equiv 1 \pmod{4}$, implying that $x+y+z$ is odd. On the other hand $(3) \; a + b + c = 2(x + y + z) = m^2 - d = m^2 - n^2 = m^2 - 25 \equiv 0 \pmod{8}$, because $m$ is odd. So $x+y+z$ is even by (3). In other words, $x+y+z$ is both odd and even, a contradiction which means that $n \neq 5$. Therefore $n=6$ and $d=6^2=36$, which inserted in (1) gives $(4) \; a^2 + b^2 + c^2 = 693$. Consequently $\sqrt{693/3} = \sqrt{231} \leq \; c < \; \sqrt{693}$, so $16 \leq c \leq 26$, implying $(5) \; 16 < \: a + b + c \: \leq 3 \cdot 26 = 78$. Moreover $a + b + c = m^2 - d = m^2 - 6^2 = m^2 - 36$, so $16 < \: m^2 - 36 \: \leq 78$, implying $8 \leq m \leq 10$. The fact that $m$ is odd gives $m=9$, hence $a^2 + b^2 + c^2 = 693$ and $a + b + c = 45$. Substituting $c$ by $45-a-b$ in the first equation result in the equation $b^2 + (a- 45)b + a^2 - 45a + 666 \:=\: 0$ which have the solution $b \:=\: \frac{45 - a \pm \sqrt{-3a^2 + 180a - 639}}{2}$. It is easy to verify that $a=12$, $b=15$ and $c=18$ is the unique solution to this problem. Conclusion: $n=6$ and $m=9$.