Find an integer $a$ such that $(x-a)(x-10)+1=(x-b)(x-c)$ for some integers $b$ and $c$

Question

Can someone help with this Olympiad question?

Find an integer $a$ such that $$(x-a) (x-10) +1$$ can be factored as $$(x-b) (x-c)$$ with $b$ and $c$ integer.

Answer 1

Hint: The discriminant $(a+10)^2-4(10a+1)=(a-10)^2-4=(a-8)(a-12)$ is a perfect square. Alternatively, show that $|b-10|=1$.

Answer 2

You want $b+c=a+10$ and $bc=10a+1$, which implies $$ bc=10(b+c-10)+1=10(b+c)-100+1 $$ hence $$ c=\frac{10b-100+1}{b-10}=10+\frac{1}{b-10} $$ Can you go on?