Proof that x^2+10x-3 is irreducible over Q

[mint grotto]

I wanted to solve this without quadratic formula, discriminate, completing the square, or rational root theorem, or Eisensteins criterion

What do you think?

[rocky berry]

idk

[old grotto]

I think it should be fine because you explain every case. With a short look to your proof i dont see mistakes.

Here some suggestions :

• irreductible is a statement for polynomials, (x^2+10x-3 for exemple) so you dont have to put =0 to make an equation. The statement would be : proof that x^2+10x-3=0 has no rationnal solutions.

• it's a very long proof, in order to help People to understand it you can add sentences at the begining of each part

[mint grotto]

Thank you!!

There are minor notation mistakes, everywhere that is says "element of Z[x]" or "element of Q[x]" is meant to be "element of irr(Z[x])" and such

[mint grotto]

Thank you for reading it! That make me super happy

Where is says let f, g=(3,1) it should say 3,-1

Also where it says gcd{fa, b}=/=1 when 3a^2/b = p| p is an integer

Is not quite right

Is should say gcd{fa, b}=/=1 or b=1 when 3a^2/b = p| p is an integer

[gloomy vessel]

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