Consider the following steps in a proof that \(\sqrt{2} + \sqrt{3}\) is irrational.
If an integer \(a\) is not divisible by 3, then \(a = 3k \pm 1\), for some integer \(k\). In both cases, \(a^2\) is one more than a multiple of 3.
Suppose that \(\sqrt{2} + \sqrt{3}\) is rational, and equal to \(\frac{a}{b}\), where \(a\) and \(b\) are positive integers with no common factor greater than one.
Then \(a^4 + b^4 = 10a^2b^2\).
So if \(a\) is divisible by 3, then \(b\) is divisible by 3.
Hence \(\sqrt{2} + \sqrt{3}\) is irrational.
Show clearly that steps 1, 3 and 4 are all valid and that the conclusion 5 follows from the previous steps of the argument.
Prove, by means of a similar method but using divisibility by 5 instead of 3, that \(\sqrt{6} + \sqrt{7}\) is irrational.
Why can divisibility by 3 not be used in this case?
Solution:
Step 1: There are only three possibilities for the remainder of \(a\) when divided by \(3\), (\(0\), \(1\), \(2\)). \(a = 3m+r\). If \(r = 0\) we are done, if \(r = 1\) take \(k = m\), and \(r=2\) take \(k=(m+1)\) and we have \(a = 3k-1\) as required.
Then \(a^2 = (3k\pm1)^2 =9k^2\pm6k+1 = 3(3k^32\pm2k)+1\) which is clearly \(1\) more than a square.
Step 3: \begin{align*}
&& \frac{a}{b} &= \sqrt{2}+\sqrt{3} \\
\Rightarrow && \frac{a^2}{b^2} &= 5+2\sqrt{6} \\
\Rightarrow && \frac{a^2}{b^2}-5 &= 2\sqrt{6} \\
\Rightarrow && 24 &= \left ( \frac{a^2}{b^2}-5 \right)^2 \\
&&&= 25 + \frac{a^4}{b^4}-10\frac{a^2}{b^2} \\
\Rightarrow && -b^4 &= a^4-10a^2b^2 \\
\Rightarrow && a^4+b^4 &= 10a^2b^2
\end{align*}
Step 4: If \(a\) is divisible by \(3\) then \(b^4 = 10a^2b^2-a^4\) is a multiple of \(3\), but if \(b\) was not a multiple of \(3\) then \(b^2\) would be \(1\) more than a multiple of \(3\) (by Step 3) and \(b^4\) would also be \(1\) more than a multiple of \(3\), and we would have a contradiction.
Step 5: Follows since either \(a,b\) are both divisible by \(3\) (contradicting Step 2), or neither is, but then \(a^2\) and \(b^2\) are both one more than a multiple of \(3\) and the RHS is one more than a multiple of \(3\) but the LHS is \(2\) more than a multiple of \(3\) which is a contradiction.
Step 1: If \(a\) is not divisible by \(5\) then \(a^2 \equiv \pm 1 \pmod{5}\)
Step 2: Suppose \(\frac{a}{b} = \sqrt{6}+\sqrt{7}\)
Step 3: \begin{align*}
&& \frac{a}{b} &= \sqrt{6}+\sqrt{7} \\
\Rightarrow && \frac{a^2}{b^2} &= 13 + 2\sqrt{42} \\
\Rightarrow && 168 &= \left (\frac{a^2}{b^2} - 13 \right)^2 \\
&&&= 169 - 26 \frac{a^2}{b^2} + \frac{a^4}{b^4} \\
\Rightarrow && a^4+b^4 &= 26a^2b^2
\end{align*}
Step 4: If \(a\) is a multiple of \(5\) then so is \(b^4\) and hence so is \(b^2\) and \(b\).
Step 5: But the left hand side is always \(2 \pmod{5}\) and the right hand side is never \(2 \pmod{5}\) contradiction.
Divisibility by \(3\) doesn't work here since mod \(3\) we can have \(a = 1, b = 1\) and have a valid solution.