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I was looking at a Mathologer video recently when he was showing how to prove Fermat's last theorem for powers of $4$. And the guy speaking went on this tangent which involved taking remainders when dividing by something. As he was speaking, he showed an equation and divided all the terms by $4$. And using some already known theory he showed how some of those terms gave only $1$ specific remainder. What I don't get is why if you divide all terms in an equation by a number, the remainders on each side of the equation are equal.

Here is what he basically meant if you don't understand what I'm saying:

$$3987^{12} + 4365^{12} = 4472^{12}.$$

And he showed that this was false because apparently and even number raised to an even number, it will always be divisible by $4$ (Can someone show we why this is true too?) And when an odd number is raised to an even number, the only remainder is $1$ (Also this one as well please). So after dividing both sides of that equation with the number $4$, the first term on the left ($3987^{12}$) gives a remainder of $1$ (because its odd raised to even) and so does the other term on the left. But the term on the right when divided by $4$ gives remainder $0$. And so he concludes that if you add the remainders on the left it gives $1+1 = 2$, and the right gives $0$. And $2$ does not equal $0$, therefore the original equation was false.

So basically summed up, my question is: Why when you divide both sides of an equation by a number, the remainders on each side of the equation will be equal?

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  • $\begingroup$ Oh btw, here's the video: youtube.com/watch?v=AO-W5aEJ3Wg $\endgroup$ – Plzhelp May 28 '18 at 16:52
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    $\begingroup$ If you apply any operation to equal quantities, you get equal results. $\endgroup$ – Lord Shark the Unknown May 28 '18 at 16:55
  • $\begingroup$ I mean, I guess so, but I'm still not satisfied. I feel like getting remainders or taking the modulus of something will give random values, and so like maybe sometimes it won't be the same? $\endgroup$ – Plzhelp May 28 '18 at 16:56
  • $\begingroup$ Like can you prove that? XD $\endgroup$ – Plzhelp May 28 '18 at 16:57
  • $\begingroup$ Let $2k$ be any even number then $(2k)^2=4k^2$, this is clearly divisible by 4. Let $2k+1$ be any odd number then $(2k+1)^2=4k^2+4k+1=4(k^2+k)+1$ clearly leaves remainder 1 upon division by 4. $\endgroup$ – user428700 May 28 '18 at 17:00
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This is important business, and I want to deal with two separate aspects of your misunderstanding.

I’ll deal only with your last sentence, “Why when you divide both sides of an equation by a number, the remainders on each side of the equation will be equal.”

The first aspect is the meaning of the mathematical symbol “$=$”. It means is. It means that you’re not talking about two numbers, the one on the left versus the the one on the right side of the equals sign. It means that you’re talking about one number, which merely has been described in two different ways. This is what justifies the standard steps that you take when solving an equation, for instance. When we take into consideration what your equation means, what you’re asking is, why the remainder upon division (by $4$ in this case) is well-defined.

And this brings us to the second thing that you’ve misunderstood, and it’s something we see frequently in our students. It’s the amazing strength of the concept of uniqueness. It’s just that if we’ve proved, somehow, that a condition is satisfied by only one number, and if we have (supposedly) two numbers that satisfy that condition, then they are in fact the same number. So from a logical condition we have derived an equality.

In the case of division, when the givens are a dividend $N$ and a divisor $d>0$, then there are a quotient $q$ and remainder $r$, such that $$N=dq+r\,,$$ satisfying $0\le r<d$. And part of the statement of the relevant theorem is that the pair of numbers $(q,r)$ so gotten is unique. And for us, it’s the uniqueness that carries the heaviest load of meaning and utility.

You have a number, it’s the number on the two sides of your equation; and you divide it by $4$; then you ask what the remainder is. Whatever it is, it’s a particular number, not “random”, as you feared in a comment.

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  • $\begingroup$ Thanks so much, I feel like I've forgotten what equality is in mathematics. Thanks a lot! $\endgroup$ – Plzhelp May 29 '18 at 12:50
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Think of it this way, any odd number can be written as $(4\times (k^2+k))+1$ so when you divide by $4$ the remainder is always one.

And also you can write an even number as $2\times a$ for any integer $a$,

so even raised to even is $(2a)^{2b}= 4^b\times a^{2b}\equiv 0\mod4 $ so remainder is always $0$

EDIT:

to answer your main question,

The sides of an equation relate both the values, like if $a=b$ there is an equivalence between them.

now consider some $x$ and you relate $x=a$dividing both sides by some number $n$ leaves a remainder.

since $a=b$ $\implies x=b$, now if you divide by the same number $n$ and get a different remainder this would mean the two numbers $a,b$ are different. So if they have the same remainder the equivalence holds. Hope I was able to answer your question.

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  • $\begingroup$ Thanks, Someone already answered this but I get it now :) $\endgroup$ – Plzhelp May 28 '18 at 17:02
  • $\begingroup$ Can someone answer like my main question tho? XD $\endgroup$ – Plzhelp May 28 '18 at 17:57

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