Recent content by Johulus

Thank you so very much :) ! You've helped a big deal. I understand everything now. I've mixed up conditions with assumption and everything but now it's all clear.

I am still not completely sure how $$ a=b $$ contradicted to $$ a \neq b \neq c $$ proves that $$ \nexists x\in\mathbb{Z}: f(x)=0 $$. I am not sure what do you mean when referring to 'opposite proposition'? That kind of bothers me. Opposite of what? I will have a look at it tomorrow. But...

1 can be divided by -1 and 1... right? Since $$(d-a)|(f(d)-f(a)) \qquad (d-b)|(f(d)-f(b)) \qquad (d-c)|(f(d)-f(c))$$ $$ (f(d)-f(a))=(f(d)-f(b))=(f(d)-f(c))=1 $$, that is: $$(d-a)|1 \qquad (d-b)|1 \qquad (d-c)|1$$ Since 1 has two divisors -1 and 1, then at least two of $$(d-a), \, (d-b), \...

I would just like to say before I write anything else that I am not that experienced with this but I would really like to figure this proof out. The first problem is that I am not that familiar with the meaning of the word lemma. But I looked for its meaning now,so I guess it's kind of clear...

I am stuck with one proof and I need some help because I don't have any idea how to proceed at this moment. The task says: If f(x) is a polynomial with integer coefficients, and if f(a)=f(b)=f(c)=-1, where a,b,c are three unequal integers, the equation f(x)=0 does not have integer solutions...