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$x\notin 19\mathbb{Z}$

If $19|x^2$ then $19|x$.

Proof by contraposition:

If $19\not|x$, then $19\not|x^2$

If $19\not|x$ then $x$ must take on one of the following forms:

$x=19k+1, x=19+2, x=19k+3,......,x=19k+18$ for some $k$

1) $x=19k+1$

$x^2=(19k+1)^2=361k^2+38k+1=19(19k^2+2k)+1$

2)$x=19k+2$

$x^2=(19k+2)^2=361k^2+76k+1=19(19k^2+4k)+1$ .

.

. . continuing this route 5)$x=19k+5$

$x^2=(19k+5)^2=361k^2+192k+25=19(19k^2+10k+1)+6$

again continuing

18)$x=19k+18$

$x^2=(19k+18)^2=361k^2+684k+324=19(19k^2+36k+17)+1$

Note how for each case, $x^2$ is not divisible by $19$ as we are always left with a remainder

$\therefore\;$ The original statement is false.

Jack's wasted life
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OLE
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    Is $19$ prime? What is the definition of prime? One of the most common/useful definitions of prime numbers is $p$ is prime iff for every $a,b$ you have $p\mid ab\implies p\mid a$ or $p\mid b$. – JMoravitz Nov 02 '16 at 02:37
  • that helps however what if my statements are contain just variables like $b^2|c^2$ then $b|c? What would work in this case? – OLE Nov 02 '16 at 03:20
  • See also both of my answers here for some conceptual proofs by descent. – Bill Dubuque Nov 02 '16 at 03:33
  • And see here for a few other proofs. – Bill Dubuque Nov 02 '16 at 03:42
  • @EduardoO. For your question in the above comment on $,b^2\mid c^2,\Rightarrow,b\mid c,$ one can use the Rational Root Test, e.g. see this answer. – Bill Dubuque Nov 02 '16 at 15:16
  • Whoops Bill I meant $b^3|c^2$ – OLE Nov 03 '16 at 01:30
  • @BillDubuque I redid the proof can you atleast let me know if it is solid? – OLE Nov 05 '16 at 23:51
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    This is still a duplicate question, and thus I'm voting to leave it closed. – apnorton Nov 06 '16 at 03:48
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    @EduardoO. Yes, you can certainly prove it by brute force checking that $,x^2\not\equiv 0\pmod{19},$ for $,x\equiv 1,2,\ldots,18.,$ But that method can't be used to give a proof for all primes, as in the linked answer. Actually you can halve your work by noting $,18\equiv -1,\ 17\equiv -2,\ldots,,$ so you only need to check $,x\equiv 1,2,\ldots,10.\ \ $ – Bill Dubuque Nov 06 '16 at 23:36

3 Answers3

1

If $19$ does not divide $x$, then $19n = x + 1$. Thus,

$$ 19 (19n^2) = x^2 + 2x + 1 \implies 19 \; \; \text{does not divide} \; \; x^2$$

ILoveMath
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  • Please correct me if I am wrong. let's suppose $x=23$. $19$ does not divide $x=23$. Then, $19n=23+1$? I don't understand. –  Nov 06 '16 at 15:32
  • That's not correct, it is not true that $,19\mid x+1$ if $,19\nmid x,,$ e.g. $,x = 1.\ $ – Bill Dubuque Nov 06 '16 at 23:34
1

Let $p \in \mathbb Z$ be a prime number, if $p|ab \Rightarrow p|a \lor p|b$

Since $p|ab \Rightarrow ab=pq$ for some $q \in \mathbb Z$. Without lose of generality suppose $p$ does not divide $a$, therefore: $GCD(p,a)=1$. By Euclid's algorithm there exist $s,t\in \mathbb Z$ such that $1=sp+ta$

$\Rightarrow b=spb+tab=spb+t(pq)=p(sb+tq)$ and since $(sb+tq)\in \mathbb Z$ by definition $p|b$

Hope this helps!

Kat
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1

let assume $19$ divides $x^2$ and does not divide $x$

Case 1: x<19

Then, $19$ cannot a prime divisor of $x$, therefore not a divisor of $x^2$ either.

Case 2: $x>19$

Then there exist non zero positive integers $k,l$ such that $x=19k+l$ where $0<l<19$. It follows $$x^2=(19k+l)^2=(19k)^2+2(19kl)+l^2$$ Since $19$ divides $x^2$ then it must divide $l^2$. However, $0<l<19$ meaning that since $19$ is prime and greater than $l$ then it does not divide $l$. Therefore, $19$ can not divide $l^2$, which would contradict that $19 $ divides $x^2$.

  • This proof is incorrect (or incomplete). What is your justification for the claim in Case 1? – Bill Dubuque Nov 06 '16 at 23:31
  • For $x=0$, it's trivial. $1\leq x <19 $ means that $x$ can take the following values $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18$. None of those values is divisible by $19$ because the latter is a prime number. Then,taking the square (or any power) of any of those numbers cannot yield a multiple of $19$ according to the Fundamental theorem of Arithmetic. –  Nov 07 '16 at 00:11
  • That brute force proof is already in the OP. – Bill Dubuque Nov 07 '16 at 00:14
  • so what is your point? What's missing in my approach? please care to share. –  Nov 07 '16 at 00:17