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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...

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

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    $\begingroup$ We need to prove: $49|a^2-5a-6$!!! $\endgroup$ – Hamid Reza Ebrahimi Jun 4 '16 at 9:43
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    $\begingroup$ I realized my mistake and edited the question before your comment. Notice that every term on the LHS is divisible by $49$, as $7|5a-2$ $\endgroup$ – Emre Jun 4 '16 at 9:44
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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 $2(a+1)+5a-2=7a$

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  • $\begingroup$ Not a big difference, but $(5a-2)+7=5(a+1)$ is a more straight-forward way for saying $7|a+1$. $\endgroup$ – Emre Jun 4 '16 at 9:45
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$$a^2-5a-6=(a+1)(a-6)$$

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

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

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  • $\begingroup$ How it proves $49|a^2-5a-6$ it gives trivial result for a=6 $\endgroup$ – Shona Jun 4 '16 at 10:25
  • $\begingroup$ @Shona, If $a\equiv6\pmod7, 7\mid (a+1)\equiv a-6$ $\endgroup$ – lab bhattacharjee Jun 4 '16 at 10:29
  • $\begingroup$ Ok thanks for explaing ☺ $\endgroup$ – Shona Jun 4 '16 at 10:52
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Hint $\ {\rm mod}\ 7\!:\ 5a\!-\!2\equiv 5(\color{#c00}{a\!+\!1})\,$ and $\,a^2\!-\!5a\!-\!6\equiv a^2\!+\!2a\!+\!1\equiv (\color{#c00}{a\!+\!1})^2$

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