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Can someone show me how to prove if $x$ divides $x+5$ then $x$ divides $5$.

For a direct proof would you have to sub out the $5$ for some integer? Why?

Are there alternate ways to do this? Maybe induction?

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  • $\begingroup$ We want to prove that there exists k2 ∈ Z such that 5 = k2x. Let k2 = k1 − 1...... how can we set k2 equal to (k1-1) .... k2 being the second integer from unpacking divisibility of x divides 5 $\endgroup$ – shibu Feb 5 '17 at 10:28
  • $\begingroup$ Hello. It seems you do not know how to accept answers. At the left side of every answer there is a check mark for you to click it. Doing so, blank check mark becomes green and that indicates that an answer is accepted. Thank you. $\endgroup$ – Juniven Feb 6 '17 at 0:03
  • $\begingroup$ i can only accept one? they all were somewhat helpful. can you answer my comment above about k2 = k1 − 1 ..... or about proving this with induction? $\endgroup$ – shibu Feb 6 '17 at 0:08
  • $\begingroup$ Yes, you can only accept one. Hope you accept mine. Let me check first your comment.--:) $\endgroup$ – Juniven Feb 6 '17 at 0:11
  • $\begingroup$ In our assumption, we have $x$ divides $x+5$. Then we can find an integer $k_1$ such that $x+5=x\cdot k_1$ and so $5=(x\cdot k_1)-x$ which means that $$5=x(k_1-1).$$ We take $k_2=k_1-1$. Then $$5=xk_2$$ where $k_2$ is now an integer. This show that $x$ is a factor of $5$, that is, $x$ divides $5$. $\endgroup$ – Juniven Feb 6 '17 at 0:17
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Result: If $a|b$ and $a|c$ then $a|(b-c)$.

Since $x|(x+5)$ and $x|x$, we get $$x|[(x+5)-x]$$ and so $x|5$.

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Write out the definitions:

$$xk=x+5\iff x(k-1)=5$$

Hence $x|5$.

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Use this ,When $ a|b ,a|c \to a|mb+nc$ $$x|x+5\\x|x\\x|m(x+5)+n(x)$$ take $m=1,n=-1$ so $$x|1(x+5)-(x) \to x|5$$

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Hint. Write $5$ as the difference of two quantities both of which are divisible by $x$.

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