We have quadratic equation $x^2-4x-21-q=0$ and we know that $q$ is a prime number. The question is to find every value of $q$ where the equation has whole solutions. I found one $q=11$ but I don't understand how to prove that no other prime will satisfy this equation. Thanks in advance!
2026-05-04 21:02:44.1777928564
Whole number roots of a polynomial equation with a prime constant term
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Solved it with a hint from Mr. Peter. Taking the $D$1$=(\frac{b}{2})^2-ac=q+25$, for the solutions to be whole, $D$1 must be a perfect square, so take some $k^2=q+25$. $$q=k^2-25$$ $$q=(k+5)(k-5)$$ For $q$ to be prime $k-5$ needs to be equal to $1$ since it is one of the two factors of a prime number. $$k-5=1$$ $$k=6$$ plug in $q=6^2-25=11$. Check that value of $11$ for $q$ yields whole roots, which it does.