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I just stumbled on a nice little proof I'd forgotten about:

Show that \( ax^2 + bx + c =0 \) has no rational solutions if \(a\), \(b\) and \(c\) are all odd.

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My proof 

If \( ax^2 + bx + c =0 \) has rational solutions, it can be written as \((px+q)(rx+s)\), with \(p\),\(q\),\(r\) and \(s\) integer.

\( a= pr\) and is odd, so \(p\) and \(r\) are both odd.

Similarly, \( c = qs\) and is odd, so \(q\) and \(s\) are both odd.

\( b \) is odd, but must equal \( ps+qr\), which is even. Contradiction.

My proof 

@icecolbeveridge (hope my first tray at a content warning works). Here's mine

Any odd square is 1 mod 8, but \(b^2-4ac\) is 5 mod 8.

Another thought 

@icecolbeveridge

Your proof relates to the method of factorising that starts "find two integers whose sum is \(b\) and whose product is \(ac\)". Mine uses the quadratic formula. Now we want one that relates to completing the square.

Another thought 

@jimsimons I think the formula is completing the square in glasses and a fake moustache ;-)

Multiply it all by \( 4a\):

4𝑎²𝑥²+4𝑎𝑏+4𝑎𝑐=0

Complete the square:

(2𝑎𝑥+𝑏)²+4𝑎𝑐−𝑏²=0

... and that only works if \(b^2 - 4ac \) is square -- and by your argument, it isn't.

Another proof 

@icecolbeveridge
along the lines of the √2 proof:

Suppose p/q is a reduced rational solution. Then clearing denominators, 𝑎𝑝²+𝑏𝑝𝑞+𝑐𝑞²=0. If exactly one of 𝑝,𝑞 are even, then this is two even terms and an odd one. If neither of them is even, then it's three odd terms. So they're both even, contradicting "reduced" fraction.

cc @jimsimons @ColinTheMathmo

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