# Show that if $\ 7|5a-2$ then $\ 49|a^2-5a-6\$

Show that if $\ 7|5a-2$ then $\ 49|a^2-5a-6\$ , ($\ a$ is positive integer)
My work:
$7|5a-2 \Rightarrow\ 49|35a-14a,49a^2 \Rightarrow\ 49|14a^2+14 \Rightarrow\ 42a^2+42a,49a^2+49a\ \Rightarrow\ 49|7a^2+7a$
And I stopped here!!
I tried to make the coefficient of $a^2$ equal to 1,but no success...

Observe that $$2(5a-2)^2-49a^2+7(5a-2)=50a^2-40a+8-49a^2=a^2-5a-6$$

• We need to prove: $49|a^2-5a-6$!!! – Hamid Reza Ebrahimi Jun 4 '16 at 9:43
• I realized my mistake and edited the question before your comment. Notice that every term on the LHS is divisible by $49$, as $7|5a-2$ – Emre Jun 4 '16 at 9:44

Hints:

$a^2-5a-6=(a+1)^2-7(a+1)$ so it is enough to prove that $7\mid a+1$

Now note that $2(a+1)+5a-2=7a$

• Not a big difference, but $(5a-2)+7=5(a+1)$ is a more straight-forward way for saying $7|a+1$. – Emre Jun 4 '16 at 9:45

$$a^2-5a-6=(a+1)(a-6)$$

As $a+1-(a-6)=7,$ $$7\mid(a+1)\iff7|(a-6)$$

Now, $$5a\equiv2\pmod7\equiv2+4\cdot7\iff a\equiv6$$

• How it proves $49|a^2-5a-6$ it gives trivial result for a=6 – Shona Jun 4 '16 at 10:25
• @Shona, If $a\equiv6\pmod7, 7\mid (a+1)\equiv a-6$ – lab bhattacharjee Jun 4 '16 at 10:29
• Ok thanks for explaing ☺ – Shona Jun 4 '16 at 10:52

Hint $\ {\rm mod}\ 7\!:\ 5a\!-\!2\equiv 5(\color{#c00}{a\!+\!1})\,$ and $\,a^2\!-\!5a\!-\!6\equiv a^2\!+\!2a\!+\!1\equiv (\color{#c00}{a\!+\!1})^2$