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Here is a solution perhaps more in the spirit of MirandaMiranda's book. Given the way the question was asked I think the point is to give a proof/computation that does not use much algebraic geometry if anything at all.

First consider the intersection of the quadrics $x_0x_3=x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$.

This is easy to deal with because one can solve the equation system: Take $x_3=\dfrac{x_1x_2}{x_0}$ and substitute it in the second equation. It easily leads to $$(x_0^2+x_1^2)(x_0^2+x_2^2)=0$$ This is the equation of two pairs of skew lines forming a $4$-gon. In other words $4$ spheres, each intersecting two others forming a cycle.

Now observe that the intersection of the quadrics $x_0x_3=x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$ is a continuous degeneration of the intersection of the quadrics $x_0x_3=2x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$. Therefore the later intersection is a compact Riemann surface $T$ (I leave it to the reader to verify that this intersection is smooth) degenerating to the above cycle of $4$ spheres. It is easy to see that then $T$ is a torus and hence its genus is $1$.

Remark The algebraic geometer's way to think about this solution is the following: The quadric $x_0x_3=\lambda x_1x_2$ is the Segre embedding of $\mathbb P^1\times \mathbb P^1\to \mathbb P^3$ given by $[a:b]\times [c:d]\mapsto [\lambda ac:ad:bc:bd]$ and then the intersection of $x_0x_1=\lambda x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$ pulls back to $\mathbb P^1\times \mathbb P^1$ as the curve defined by the equation $$\lambda^2a^2c^2+a^2d^2+b^2c^2+b^2d^2=0.$$ Now this defines a divisor of degree $(2,2)$ on $\mathbb P^1\times \mathbb P^1$ which can be represented (choosing $\lambda=1$ for instance) by two pairs of lines as described above. If one knows about the behaviour of the (arithmetic) genus in flat families, then everything claimed above is clear.

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Here is yet another a solution , perhaps more in the spirit of Miranda. Given the way the question was asked I think the point is to give a proof/computation that does not use much algebraic geometry if anything at all.

First consider the intersection of the quadrics $x_0x_3=x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$.

This is easy to deal with because one can solve the equation system: Take $x_3=\dfrac{x_1x_2}{x_0}$ and substitute it in the second equation. It easily leads to $$(x_0^2+x_1^2)(x_0^2+x_2^2)=0$$ This is the equation of two pairs of skew lines forming a $4$-gon. In other words $4$ spheres, each intersecting two others forming a cycle.

Now observe that the intersection of the quadrics $x_0x_3=x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$ is a continuous degeneration of the intersection of the quadrics $x_0x_3=2x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$. Therefore the later intersection is a compact Riemann surface $T$ (I leave it to the reader to verify that this intersection is smooth) degenerating to the above cycle of $4$ spheres. It is easy to see that then $T$ is a torus and hence its genus is $1$.

Remark The algebraic geometer's way to think about this solution is the following: The quadric $x_0x_3=\lambda x_1x_2$ is the Segre embedding of $\mathbb P^1\times \mathbb P^1\to \mathbb P^3$ given by $[a:b]\times [c:d]\mapsto [\lambda ac:ad:bc:bd]$ and then the intersection of $x_0x_1=\lambda x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$ pulls back to $\mathbb P^1\times \mathbb P^1$ as the curve defined by the equation $$\lambda^2a^2c^2+a^2d^2+b^2c^2+b^2d^2=0.$$ Now this defines a divisor of degree $(2,2)$ on $\mathbb P^1\times \mathbb P^1$ which can be represented (choosing $\lambda=1$ for instance) by two pairs of lines as described above. If one knows about the behaviour of the (arithmetic) genus in flat families, then everything claimed above is clear.

4 added 2 characters in body

Here is yet another solution, perhaps more in the spirit of Miranda. I think the point is to give a proof/computation that does not use much algebraic geometry if anything at all.

First consider the intersection of the quadrics $x_0x_3=x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$.

This is easy to deal with because one can solve the equation system: Take $x_3=\dfrac{x_1x_2}{x_0}$ and substitute it in the second equation. It easily leads to $$(x_0^2+x_1^2)(x_0^2+x_2^2)=0$$ This is the equation of two pairs of skew lines forming a $4$-gon. In other words $4$ spheres, each intersecting two others forming a cycle.

Now observe that the intersection of the quadrics $x_0x_3=x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$ is a continuous degeneration of the intersection of the quadrics $x_0x_3=2x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$. Therefore the later intersection is a compact Riemann surface $T$ (I leave it to the reader to verify that this intersection is smooth) degenerating to the above cycle of $4$ spheres. It is easy to see that then $T$ is a torus and hence its genus is $1$.

Remark The algebraic geometer's way to think about this solution is the following: The quadric $x_0x_3=\lambda x_1x_2$ is the Segre embedding of $\mathbb P^1\times \mathbb P^1\to \mathbb P^3$ given by $[a:b]\times [c:d]\mapsto [\lambda ac:ad:bc:bd]$ and then the intersection of $x_0x_1=\lambda x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$ pulls back to $\mathbb P^1\times \mathbb P^1$ as the curve defined by the equation $$\lambda^2a^2c^2+a^2d^2+b^2c^2+b^2d^2=0.$$ Now this defines a divisor of degree $(2,2)$ on $\mathbb P^1\times \mathbb P^1$ which can be represented (choosing $\lambda=1$ for instance) by two pairs of lines as described above. If one knows about the behaviour of the (arithmetic) genus in flat families, then everything claimed above is clear.

3 Modified the presentation to avoid confusing argument.; added 2 characters in body; edited body; edited body
2 removed confusing statement about triangulations
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