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Have to prove - If $a | (b+c) $ and $a | (b-c) $, then $a | b $ and $a | c $ Don't really know, which approach to take. Thought maybe to break this statement on parts, and prove:

  1. If $a | (b+c) $, then $a | b $ and $a | c $
  2. If $a | (b-c) $, then $a | b $ and $a | c $ But don't know if it is going to be valid. Can someone suggest which approach to take?
Sebastiano
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wintermute
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    It isn't true. Take $(a,b,c)=(2,3,1)$ for example. – lulu Feb 04 '23 at 19:44
  • But does the wrong example proves, that the statement is wrong in every case? Excuse me, I'm asking genuinely, not that fluent. I thought the idea, is to prove it formally – wintermute Feb 04 '23 at 19:46
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    If a divides 2 numbers, it must also divide their difference and sum, try and build onto that – Fix Feb 04 '23 at 19:46
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    If $a$ is odd the claim is true, otherwise it's false in general, as per the counterexample given by lulu. – Bruno B Feb 04 '23 at 19:53
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    The wrong example shows that the statement "For every $a,b,c$, if $a|b+c$ and $a|b-c$, then $a|b$ and $a|c$" is false. – Sambo Feb 04 '23 at 19:54
  • Another set of examples is to let $c = b$ and $a=2b$. – Cameron Williams Feb 04 '23 at 19:56
  • See the linked dupe a the correct statement and methods of proof for such problems, e.g. the methods there derive $,a\mid 2b,2c,$ so $,a\mid (2b,2c) = 2(b,c)\ \ $ – Bill Dubuque Feb 04 '23 at 20:01
  • @BrunoB But how to prove, that when a is odd the claim is true. How did you see that? Would it be valid, to make an assumption from the beginning : Let's suppose a is odd, then ... – wintermute Feb 04 '23 at 20:03
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    If you miscopied the problem then let me know the correct statement so we can make the dupe links more precise (most all problems of this sort are by now dupes). – Bill Dubuque Feb 04 '23 at 20:07
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    Hopefully you have at least understood why $a$ divides $2b$ and $2c$ (with Bill's link or with this one). Then, the reason why $a$ odd makes the claim true is simply because if $a$ divides $2b$, then we have $ka = 2b$, but since $a$ is odd, $k$ as to be even, otherwise the odd number $ka$ would be equal to the even number $2b$, thus $k = 2k'$. Therefore $2k'a = 2b$, and so $k'a = b$, hence $a$ divides $b$, and then it's the same for $c$. (Might probably count as an answer of sorts but the question is closed so it'll be a comment instead) – Bruno B Feb 04 '23 at 20:31
  • @BrunoB What i've done was - a | ( b + c) and a | ( b - c ), then b + c = am, and b - c = an . So b = am - c and b=an + c ; am - c = an + c , a ( m - n ) = 2c ; ap = 2c, where p is an integer and equals to ( m - n ). That's what you've said in the first part of the comment. What i don't get, is the calm but since a is odd, k as to be even. Why the a is odd. Because a can be even also. Unless, at the beginning of the proof, this claim is set. That a is odd. However, i don't understand, why is it valid to simply tell, that a is odd. – wintermute Feb 04 '23 at 20:51
  • While i don't show, that this claim doesn't hold for a to be even. However, in the expression ap = 2c, a can be even, and p can be odd. Or both to be even. Sorry for me being stubborn, seems i'm lacking knowledge, and don't understand. – wintermute Feb 04 '23 at 20:54
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    I decided to set $a$ odd because that way the claim is true AND because there already are counterexamples provided for the general claim when $a$ wasn't of fixed parity. To prove your general claim false, it is not necessary to prove that for all $(a,b,c)$ with $a$ even the claim doesn't hold, because it'll still hold for some of those triplets, instead it suffices to only check that one such triplet makes the claim false. – Bruno B Feb 04 '23 at 20:58

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