For all integers b, c, and d, if x is a rational numbe such that x^2 + bx + c =

Question

For all integers b, c, and d, if x is a rational numbe such that x^2 + bx + c = d, then x is an integer.

Not sure how I would go about starting to figure out whether or not this true or false.

The hints that i was given are, "Are all the quantifiers given explicitly? It is okay, but not necessary to sue the quadratic formula."

Explanation / Answer

So we are given x^2+bx+c=d iff x^2+bx+(c-d)=0 here let c-d = h Then x^2+bx+h=0 where b,h are integers Say x is rational x=p/q(p divided by q) say q>0 and gcd(p,q)=1 Hence (p/q)^2 + b(p/q) + h = 0 iff p^2 + bpq + hq^2=0 iff p^2 = -hq^2 - bpq So q|RHS hence q|LHS hence q|p^2 but gcd(p,q) = gcd(p,q^2)=1 Hence q = 1 as q|p^2 and gcd(q,p^2)=1 Hence x = p Hence x is integer Hence proved

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