It same as proving, that there are not integer numbers $a,b,c,d$ with $a+b+c=0$ and $a^2+b^2+c^2=d^2$ with $d \neq 0$ $d^2=a^2+b^2+c^2=2(a^2+b^2+ab)$ But $deg_2(2(a^2+b^2+ab))$ is odd, and $deg_2(d^2)$ is even

RagvaloD wrote: It same as proving, that there are not integer numbers $a,b,c,d$ with $a+b+c=0$ and $a^2+b^2+c^2=d^2$ I don't think so. Your system has a solution $(a,b,c,d)=(0,0,0,0)$ whereas the original problem has not.

Yes, also $d \neq 0$ is needed

let $w$ be the least integer such that $wx,wy,wz\in\mathbb{Z}$. We'll say $a,b,c=wx,wy,wz$ respectively. We will also say that $x,y,z\not=0$ It's then clear to see that $\gcd(a,b,c)=1$. We then have: $$a^2+b^2+c^2=100w^2$$ $$a^2+b^2=100w^2-c^2$$ $$a^2+b^2=(10w-c)(10w+c)$$ Since $c$ and $w$ are integers, RHS cannot be congruent to $1$ mod $4$. Furthermore, since LHS is a sum of two squares, it cannot equal $3$ mod $4$. This means that they both must be even, implying that $c$ is even. Since the problem is symmetric, the same argument can be applied to $a$ and $b$. This contradicts $\gcd(a,b,c)=1$, therefore at least one of $x$ $y$ or $z$ is zero. WLOG we say that $z=0$ We then have $$x^2+y^2=100$$ $$x+y=0$$ Solving this, it's clear to see that $x$ and $y$ are irrational, concluding the proof.

Let's introduce the elementary symmetric functions of three variables which we denote by $\sigma_{i}$.. Then the given situation is interpreted as follows: We've $\sigma_{1}=0$ and $\sigma_{1}^{2}-\sigma_{2}=50 \Leftrightarrow \sigma_{2}=-50$ Therefore it follows that $x,y,z$ are zeros cubic $t^{3}-50t+k=0$ where we've assumed that $\sigma_{3}=xyz=k \in \mathbb{Q}$(suppose) (Note: If $xyz$ is non rational , we're done as then the cubic will have no real zeros) If $r$ is a rational zero of the family of cubics of the above form, then by a fundamental and deep result it must be an integer, then we get $r(50-r^2)=k$.. From this we've the following cases: Case 1 $(r,k)=(1,49)$ We can verify that the resulting cubic $t^3 -50t +49=(t-1)(t^2 +t -49)$ has one real and two complex numbers as zeros Case 2 $(r,k)=(7,7)$ We can similarly verify that the cubic $t^2 -50t +7=(t-7)(t^2 +7t -1)$ has one real and two complex numbers as zeros.. Hence, there don't exist rational numbers satisfying the given conditions.. QED

Is there a way to generalize this for 4 numbers $a,b,c,d$ or even more?

What do you want to replace the 100 with? If you have 4 integers $a,b,c,d$ with $a+b+c+d=0$, then $a^2+b^2+c^2+d^2=s$ reduces to $(a+b)^2+(a+c)^2+(b+c)^2=s$, which has or has no solutions according to the three squares theorem, assuming $s$ is even. EDIT: But it is not impossible to generalize, see https://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=249323&start=0&lps=1822053#v1822053