Perfect squares relantionship

Question

Find all natural numbers $\overline{xyzt}=10^3x+10^2y+10z+t$ who satisfies the following condition $$ \sqrt{x}+2\sqrt{y}+3\sqrt{\overline{zy}}+4\sqrt{z}=t^2. $$

Answer 1

I believe it is true, though I don't know whether there is any simple proof of this, that no linear combination of irrational square roots can be rational. In that case, each of the square roots in your condition must be integral. In particular, $\sqrt{y}, \sqrt{\overline{zy}}$ and $\sqrt{z}$ must all be integers. But this limits $\,y,z \in \{0,1,4,9\},\,$ so either $\,z=0\,$ or $\,\overline{zy}=49\,$.

If $\,z=0\,$, the condition simplifies to $\,\sqrt{x}+5\sqrt{y}=t^2.\,$ Since $\,0 \le \sqrt{n} \le 3\,$ for single-digit $n$'s, we have that $\,0 \le t^2 \le 18,\,$ and excluding the trivial $\,x=y=z=t=0\,$ case, that leaves $\,t^2 \in \{1,4,9,16\}.\,$ A little trial and error gives the only solutions to be $\,\overline{xyzt}=1001\,$ and $\,1904.\,$

If $\,\overline{zy}=49,\,$ the condition simplifies to $\,\sqrt{x}+35=t^2,\,$ and the bounds on $\,\sqrt{x}\,$ force $\,t^2=36,\,$ i.e. $\,x=1,\,$ which yields the single solution $\,\overline{xyzt}=1946.\,$

Thus the only solutions are

$$\overline{xyzt}=\boxed{1001,1904,1946}$$

Answer 2

$$x=y=z=9$$ and $$\overline{zy}=81$$ gives $48$ so $$t<7$$ Further $$x=y=z=1$$ and $$\sqrt{\overline{zy}}=\sqrt{16}=4$$ gives $$t\geq 5$$ so we have $$5\le t\le 6$$, we have only two cases to consider $$t=5,t=6$$