Prove that $\sqrt{7}+\sqrt{3}$ is irrational Is there a method by which we can prove that $$\sqrt{3}+\sqrt{7}$$ is irrational. It's obviously an irrational number, but I want to prove that mathematically.
 A: Consider
\begin{align}
&a = \sqrt{7}+\sqrt{3}&&(1)\\
&a(\sqrt{7}-\sqrt{3})=(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})&&(2)\\
&\sqrt{7}-\sqrt{3}=\frac{4}{a}&&(3)\\
&2\sqrt{7}=a+\frac{4}{a}&&\text{summing the reverse of (1) with (3)}
\end{align}
If $a$ is rational, then also $\sqrt{7}$ is rational.
A: If $a:= \sqrt{3} + \sqrt{7}$ were rational, so would $\frac{a^2 - 10}{2}$ be, which equals $\sqrt{21}$. So suppose $\frac{p}{q} = \sqrt{21}$, then $p^2 = 21q^2$. The right hand side has an odd number of prime factors $3$ (a square always has an even number (including $0$) of any prime), and the left hand side has an even number. This is a contradiction. 
A: Notice that $(\sqrt 3+\sqrt 7)^2=10+2\sqrt{21}$, so  that $((\sqrt 3+\sqrt 7)^2-10)^2=84$. This means that the number $\alpha=\sqrt3+\sqrt7$ is a root of the polynomial $$f(X)=(X^2-10)^2-84=x^4-20 x^2+16.$$
You can now use the rational root test to show that $f$ does not have any rational roots: indeed, it follows from that result that all rational solutions are actually integers which divide $16$, and you can easily check that no integer dividing $16$ is a root of $f$.
