1

By assumption $y = x + km$ ; some $k \in \mathbb{Z}$.

w.t.s $ay^2 + by + c = (ax^2 + bx + c$) + $k'm$ ; some $k' \in \mathbb{Z}$

If we substitute our assumption in for $y$ we have

$a(x + km)^2 + b(x + km) + c$

= $a(x^2 + 2xkm + k^2m^2) + bx + c + (bk)m$

= $ax^2 + 2xkma + ak^2m^2 + bx + c + (bk)m$

= $ax^2 + bx + c + (2xka + k^2ma + bk)m$

Lastly, set $k' = 2xka + k^2ma + bk$ and we have

$ay^2 + by + c = (ax^2 + bx + c$) + $k'm$

does this work? heavily realying just on closure of $\mathbb{Z}$ under + and $\cdot$

(I'm fairly new to number theory)

Also, if this has been asked please just refer me to the Paige, I couldn't finds it. I apologize in advance if it has been.

trancelocation
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homosapien
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4 Answers4

2

Your proof is correct, but in a sense it misses the point of the exercise. Here is a better proof:

Assume that $x \equiv y \mod m$. Then, $x^2 \equiv y^2 \mod m$ (since we can take a congruence to the $2$-nd power). Thus, $ax^2 \equiv ay^2 \mod m$ (since we can multiply a congruence by an integer constant -- in this case, $a$). Also, $ax \equiv ay \mod m$ (here, we have multiplied the original congruence by $a$). Finally, $a \equiv a \mod m$. Now, adding the last three congruences together, we obtain $ax^2 + bx + c \equiv ay^2 + by + c \mod m$. Done.

Why do I call this proof "better"? Well, try applying your idea to show that $\left(ax^4+b\right)\left(c-dx^7\left(x+e\right)\right) \equiv \left(ay^4+b\right)\left(c-dy^7\left(y+e\right)\right) \mod m$ instead :) My proof, on the other hand, would still apply here, except that you would have to multiply and subtract congruences as well. The general principle at hand here is the principle of substitutivity for congruences (see §2.2.7 of my Notes on the combinatorial fundamentals of algebra, 10th of January 2019, for example).

darij grinberg
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  • thanks for this, could you look over my solution? is that what you were going for? I actually like yours more come to think of it, mine ( my initial solution) is too elementary and its just basic algebra manipulation (too boring). – homosapien Feb 08 '19 at 18:06
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    @HossienSahebjame: Yes, both your solutions are correct (and the second one is basically the same as mine). – darij grinberg Feb 08 '19 at 18:21
1

You may just write

$$a(x^2-y^2) + b(x-y) \equiv (x-y)(a(x+y) +b) \equiv 0 \mod m$$

trancelocation
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  • LOL okay what? I would never come up with this on my own. Is my way correct? please elaborate on yours, I mean I see how the first $\equiv$ holds, but why does the second and how does this show what Im showing? is it because $(a(x + y) + b$ is just an integer so since $m$ divides $(x - y)$ it divides $(x - y)$ times and integer? – homosapien Feb 07 '19 at 04:24
  • you said $x\equivelant y modm$ – Seth Feb 07 '19 at 04:27
  • @HossienSahebjame You may use directly that $x \equiv y \mod m$ means $x-y = k\cdot m$ for some $k \in \mathbb{Z}$. – trancelocation Feb 07 '19 at 04:27
  • ^thats what i was trying to say – Seth Feb 07 '19 at 04:28
  • Im so sorry lol I JUST got your method and it is clever. so my way also works? – homosapien Feb 07 '19 at 04:28
  • @HossienSahebjame I am checking it right now. just give me a second. – trancelocation Feb 07 '19 at 04:28
  • @HossienSahebjame Yes, your way looks fine. There was just a little typo I edited right now. There was a factor $a$ too much while expanding $(x+km)^2$ – trancelocation Feb 07 '19 at 04:32
  • Oh awesome thanks so much!! also thank you for yours!!! I love seeing as many ways as possible and it helped me think why each step was true and how/why to apply my assumption. – homosapien Feb 07 '19 at 04:35
  • You are welcome :-) – trancelocation Feb 07 '19 at 04:37
1

Prove things once. Then assume they are written in stone.

Prove once that if $a\equiv a'\pmod n$ and $b\equiv b'\pmod n$ then $a \pm b\equiv a' \pm b'\pmod n$.

Then prove once that if $a \equiv a' \pmod n$ and $b\equiv b'\pmod n$ then $ab \equiv a'b'\pmod n$.

Then observe once by induction that if $a \equiv a'$ then $a^m \equiv a'^m \pmod n$ for all positive powers $m$.

And observe once for any integer $d$ that $d \equiv d \pmod n$ for any integer $n$.

Then this is ... just observation.

$x \equiv y \pmod m$ so $x^2 \equiv y^2$ and as $a \equiv a \pmod m$ then $ax^2 \equiv ay^2$ and as $b \equiv b$ and $c \equiv c$ then $bx \equiv by$ and $ax^2 + bx + c \equiv ax^2 + bx + c\pmod n$.

And that IS a complete proof.

=====

To prove these things ONCE:

$c - c = 0$ and $0 = 0*n$ for all integers $n$ so $n|c-c$ for all $n$ so

$c \equiv c \pmod n$ for any integers $c, n$.

If $n|a$ then there is an integer $k$ so that $a = kn$ and if $n|b$ then there is a $j$ so that $b = jn$ so $a\pm b = kn \pm jn = (k\pm j)n$. $k\pm j$ are both integers so

If $n|a$ and $n|b$ then $n| a\pm b$.

If $a\equiv a' \pmod n$ and $b\equiv b' \pmod n$ then $n|a-a'$ and $n|b-b'$ so $n|(a-a') \pm (b - b')=(a\pm b)-(a' \pm b')$ so

If $a \equiv a'\pmod n$ and $b \equiv b'\pmod n$ then $a\pm b \equiv a' \pm b'\pmod n$.

If $a\equiv a' \pmod n$ then there is a $k$ so that $a = a' + kn$ and if $b\equiv b' \pmod n$ then there is a $j$ so that $b = b' + jn$ so $ab = a'b' + ajn + bkn + jkn^2 = a'b' + n(aj + bk + jkn)$ and as $aj + bk + jkn$ is an integer then

If $a \equiv a'\pmod n$ and $b \equiv b'\pmod n$ then $ab \equiv a'b' \pmod n$.

If we have $a \equiv a' \pmod n$ and we have $a^k \equiv a'^k \pmod n$ for $k = 1$. If we assume that for some $k$ that $a^k \equiv a'^k$ then we have $a^{k+1} = a^ka \equiv a'^ka'= a'^{k+1} \pmod n$

So by induction $a^m \equiv a'^m\pmod n$ for all positive integer powers $m$.

....

And that's it. We have proven everything we need once. We will NEVER need to prove them again.

fleablood
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  • you went way too much man and I already got it see look – homosapien Feb 08 '19 at 14:44
  • Still the advice is: Math is tools. You buy the tools by proving them a first time. Once you buy the tools... use them. You don't need to build an house by whittling trees with a penknife when you've already had a lumber mill deliver three thousand pounds of wooden planks on your yard. – fleablood Feb 08 '19 at 17:04
  • bro I know this, I have a degree in pure math from ucla, I just have never taken number theory needed proof verification. Not a crash course on proof writing man. thanks tho. – homosapien Feb 08 '19 at 18:04
0

I got it myself.

Since $x \equiv y$ (mod $m$), by property of $\equiv$ one has

$x^2 \equiv y^2$ (mod $m$)

By scalar property one has

$ax^2 \equiv ay^2$ (mod $m$)

Also by scalar and our assumption one has

$bx \equiv by$ (mod $m$)

By additive property we have

$ax^2 + bx \equiv ay^2 + by$ (mod $m$) (**)

Note by using our assumption we also have

$x + c \equiv y + c$ (mod $m$)

(Because $m \vert (x - y) \Rightarrow m \vert x + c - y - c$), applying this to (**) and get

$ax^2 + bx + c \equiv ay^2 + by + c$ (mod $m$) and we are done.

homosapien
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