divisibility relation $a|b^2 + 10c.$

Question

Use divisibility relation to show that for all integer $a$, $b$, $c$, $a \ne 0$ counts if $a|b$ and $a|c$ then $a|b^2 + 10c$.

Use direct proof.

Ok,

$a|6$ then there is integer $k$. $$a*k=6,$$ $$a*k1=6 , $$ $$a*k2=c$$ $$ b^2+10c=a*k3$$

then? :/

Answer 1

Suppose that $a\mid b$ and $a\mid c$. By definition this means that $$b=ax\quad\hbox{and}\quad c=ay$$ for some integers $x$ and $y$. Therefore $$b^2+10c=(ax)^2+10(ay)=a(ax^2+10y)\ .$$ But $ax^2+10y$ is an integer (as it is a sum of products of integers): let's call it $z$. Therefore $b^2+10c=az$, where $z$ is an integer. By definition this means $$a\mid b^2+10c\ .$$

Answer 2

it is staightly follow as a|b imply a|b^2 and a| c imply a| 10c and further " as we know if x|y and x|z then x| x+z " by using this we can directly conclude that a | b^2 + 10c.