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Given positive integers $n, r$, with $p$ a prime, and $n^2/r^2 = p$, prove that $n$ is divisible by $p$. Use proof by contradiction (I am sure there are easier ways to prove this, but this is what the book requests, since the chapter is exercises on proof by contradiction).

I started off by assuming that $n$ is not divisible by $p$, therefore:

$n = ap + b$ for a non-negative integer $a$ and a positive integer $b$.

$n^2 = pr^2$, so $(ap+b)^2 = pr^2$

I expanded this out to get a quadratic polynomial in $b$:

$b^2 + 2apb + a^2p^2-r^2p = 0$

Using the quadratic formula, I found:

$b = -ap \pm r\sqrt{p}$

Since $a$, $r$, and $p$ are integers, and $p$ is prime, then both of these values of $b$ would therefore be irrational, violating the condition that $b$ must be a positive integer.

I'm not sure if this proof works or not. It doesn't feel very elegant (compared to some of the examples in the book). Could you please critique it?

Tumbleweed53
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  • the relationship $n^2 = p r^2$ is impossible in nonzero integers. For this reason, I like to switch to $$ n^2 \equiv p r^2 \pmod {p^2} $$ – Will Jagy Dec 19 '20 at 02:42
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    Given date in the question implies that $\sqrt{p}=\dfrac{n}{r}$ is rational. Do you know any such prime? – Bumblebee Dec 19 '20 at 03:24
  • The question is wrong. For no $n,r\ne0,p$ is $n^2=r^2p$. The exponent of $p$ is odd on the RHS and even on the LHS. – Shubham Johri Dec 19 '20 at 03:39
  • What I think is since the 'if' condition is false , the whole statement is true. – Infinity_hunter Dec 19 '20 at 03:43
  • I checked the text, which is An Introduction to Mathematical Reasoning by Iglewicz, Stoyle, and the question is stated as: if $n^2/r^2 = p$ where $p$ is a prime number, then $n$ is divisible by $p$. $n,r$ are natural numbers. What am I missing? Or is this question in error? @Bumblebee – Tumbleweed53 Dec 19 '20 at 04:16
  • Almost surely it is the first step in a sequence of exercises culminating in the common proof (by contradiction) that $\sqrt p,$ is irrational. – Bill Dubuque Dec 19 '20 at 08:57
  • $,p\mid n^2\Rightarrow p\mid n,$ is a special case of Euclid's Lemma. It's a bit unnatural to prove this Lemma by contradiction, but if you must then you can do so by asusming $,p\nmid n,$ then comparing unique prime factorizations, as in the dupe link. But this amounts to assuming $,p\nmid n,,$ then proving $,p\mid n,,$ so this use of contradiction is unnecessary since we already have a proof that $,p\mid n.,$ – Bill Dubuque Dec 19 '20 at 09:12
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    This is analogous to the unnecessary use of contrsdiction in the common proof that there are infinitely many primes (despite many claims to the contrary, Euclid did not prove this by contradiction). – Bill Dubuque Dec 19 '20 at 09:12

2 Answers2

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You proof does not work because you did not use the fact that $n$ is not divisible by $p$ (you could have $b=0$ for instance). Also, when defining $b$, I think it would be more convenient to say that $n=ap+b$ is the euclidean division of $n$ by $p$ (this also implies that $0<b<p$). What is wrong in you proof is that the initial equation is false. Indeed you found a contradiction without using the fact that $n$ is not divisible by $p$ so the initial hypothesis is false, that is $n^2=pr^2$. You can see that by writing $2v_p(n)=1+2v_p(r)$ which is not possible because the LHS is even and the RHS is odd.

Tuvasbien
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  • I am not following your quadratic formula solution. I reworked the algebra and get what I originally stated. I must be missing something obvious – Tumbleweed53 Dec 19 '20 at 02:54
  • You're right, I'm a bit tired ahah, I rectified my answer and your quadratic formula is correct. – Tuvasbien Dec 19 '20 at 03:00
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Your proof is really very overcomplicated. It's also missing something: note that you aren't assuming that $b < p$ anywhere. What goes wrong if I choose $b = p = 2, a = 1, n = 4$?

For a much easier solution: Multiply by $r^2$ to obtain $n^2 = pr^2$. Apply the definition of "prime".

user3482749
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  • Yes it is overcomplicated, can you prove it more simply by contradiction as is stated in the problem statement? I don't see a simpler one but it feels one must exist. Agreed I did miss the assumption that $b < p$, luckily it still works with that added. – Tumbleweed53 Dec 19 '20 at 02:52
  • The question is wrong. For no $n,r\ne0,p$ is $n^2=r^2p$ true. The exponent of $p$ is odd on the RHS. – Shubham Johri Dec 19 '20 at 03:38
  • @ShubhamJohri It's not "wrong". Rather it's part of a common proof by contradiction that $\sqrt p$ is irrational, see my comment on the question. – Bill Dubuque Dec 19 '20 at 09:00
  • @Tumbleweed53 Take the proof that I hinted at, wrap it in a dummy contradiction argument.

    Just adding that assumption doesn't help: you aren't using it anywhere.

     – user3482749 Dec 19 '20 at 11:01