<s>[INST] If the coefficients of a quadratic equation are odd numbers, show that it cannot have rational roots If the coefficients of a quadratic equation $$ax^2+bx+c=0$$ are all odd numbers, show that the equation will not have rational solutions.
I am also not sure if I should consider $c$ as a coefficient of $x^0$, suppose if I take that $c$ is also odd,
then $$b^2-4ac $$ will be odd. But that $-b$(odd), in the quadratic formula will cancel out the oddness of $\sqrt{b^2-4ac}$, in case if it is a perfect square. If it's not a perfect square then the root is irrational.
If I take $c$ to be even, even then the  same argument runs but we noticed that when we take $c$ odd, we get that when discriminant is perfect square, so that means the question asks for $c$ not to be an coefficient.
Final question: Is this right to take $c$ as one of the coefficients of the equation $ax^2+bx=c=0$?
 [/INST] Take $a=1,b=3,c=2$ to get the rational solutions $-2,-1$. So the statement is false unless $c$ is also required to be odd.
Now consider squares modulo $8$. Any odd number has the form $8n+1$, $8n+3$, $8n+5$, or $8n+7$ (these are abbreviated as $\equiv1,3,5,7\bmod8$). So an odd number squared is
$$1^2=1$$
$$3^2=9=8\cdot1+1\equiv1$$
$$5^2=25=8\cdot3+1\equiv1$$
$$7^2=49=8\cdot6+1\equiv1.$$
And any odd number times $4$ is
$$4\cdot1=4$$
$$4\cdot3=12=8\cdot1+4\equiv4$$
$$4\cdot5=20=8\cdot2+4\equiv4$$
$$4\cdot7=28=8\cdot3+4\equiv4.$$
Therefore, if $a,b,c$ are all odd, then $ac$ is also odd, and
$$b^2-4ac\equiv1-4=-3=8\cdot(-1)+5\equiv5\not\equiv1$$
so $b^2-4ac$ cannot be a square.
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