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I am very new to Number Theory, I understood the basic concepts and theorems but what I am struggling at is that I am not able to apply them on questions, and basically I am not able to process the problem in 'number theory' theorems rather I am thinking from an absolute manner.

I have solved around 15-20 divisibility problems of Number Theory and I still have no idea how you deal with congruences in a real problem, for example:

Prove that for positive integer $ n$ we have $169| 3^{3n+3}-26n-27$.

My approach to this problem was by simplifying the expression as much as I can but it involved congruences, which didn't even strike me. Even when I am reading congruence questions' answers, I am not able to simply understand

Now I have a number of questions:

  • first, does it take time to develop that 'intuition' while solving problems?
  • second, can elementary number theory be mastered through sheer practice?
  • third, is there a good resource to study and practice congruences?
Arturo Magidin
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Patrick Schick
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  • I doubt anyone here can tell if you need a mentor, but practice is essential. For modular arithmetic, we saw $a \equiv b \bmod m$ when $m \mid (a-b)$ and divisibility by $m$ is all about showing $a \equiv 0 \bmod m$. What is special about modular arithmetic is the congruences stay congruences under addition and multiplication of congruent numbers. You are asking about $3^{3n+3} - 26n - 27 \equiv 0 \bmod 13^2$. Why not do some algebra and rewrite that as $27^{n+1} - 26n -27 \equiv 0 \bmod 13^2$? I suggest writing $27^{n+1}$ as $(1 + 2 \cdot 13)^{n+1}$ and using the binomial theorem. – KCd Nov 16 '20 at 06:10
  • When you are working modulo $m$, a crucial thing to remember is that $m \equiv 0 \bmod m$, so all multiples of $m$ can be ignored. For example, when you expand $(1 + 2 \cdot 13)^5 \bmod 13^2$, you can expand by the binomial theorem before you reach terms with $13^2$ in them since multiples of $13^2$ are $0 \bmod 13^2$. So $(1 + 2\cdot 13)^5 \equiv 1 + 5(2 \cdot 13) \bmod 13^2$. – KCd Nov 16 '20 at 06:12
  • okay so modular arithmetic is basically simplifying the question into a simple congruence and then operate over that to come to a solution? That makes sense to be honest. SO do you suggest reading more about congruence theorems or practicing more divisibility questions using congruence? – Patrick Schick Nov 16 '20 at 06:17
  • Yes, it does take time. Yes, Ele. Num. Theory can be mastered as long as you treat it like a foreign language, stretching your intuition, working to think in terms of the foreign language, and finding a book (or books) with many problems to work through. mathSE is an excellent resource re problem assistance. As an example re developed instincts, my first thought when I saw the problem, was that since $13 | (3^3 -26 - 27)$ the real challenge will be showing that its quotient is also divisible by $13,$ so that the expression is divisible by $(13)^2.$ – user2661923 Nov 16 '20 at 06:36
  • Congruences behave in many (but not all) ways like equations, which is partly why the notation is so technically convenient. In particular, rewrite $27^{n+1} - 26n - 27 \equiv 0 \bmod 13^2$ as $27^{n+1} \equiv 27 + 26n \bmod 13^2$, or $(1 + 2 \cdot 13)^{n+1} \equiv 27 + 26n \bmod 13^2$. Expand the left side with the binomial theorem, keeping track of terms modulo $13^2$, and check you get the right side. – KCd Nov 16 '20 at 06:45
  • You can look at this thread for a few techniques and several examples. If you view the right margin (when in that thread) you will find scores of other threads linked to it for more training material. Caveat: I am not saying that the thread would be optimized pedagogically, but it does explain diverse techniques even though it is limited in scope to remainders of "large powers". – Jyrki Lahtonen Nov 16 '20 at 07:07
  • If you study this answer in the linked dupe you will see that this divisibility arises immediately from the first couple terms of the Binomial Theorem - which is a frequently applied method for problems like this. The more you practice solving problems the better you will gain intuition on these common problem solving techniques. – Bill Dubuque Nov 17 '20 at 10:14

3 Answers3

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Note that $3^{3n+3}=(26+1)^{n+1}$

Can you use binomial expansion from here?

Hint

$3^{3n+3}=(26+1)^{n+1} \\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }={{n+1} \choose {n+1}}(26)^{n+1}+{{n+1} \choose {n}}(26)^{n}+\dots+{{n+1} \choose {2}}(26)^{2}+{{n+1} \choose {1}}(26)+{{n+1} \choose {0}} \\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }={{n+1} \choose {n+1}}(26)^{n+1}+{{n+1} \choose {n}}(26)^{n}+\dots+{{n+1} \choose {2}}(26)^{2}+(n+1)(26)+1 \\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }=\color{green}{{{n+1} \choose {n+1}}(26)^{n+1}+{{n+1} \choose {n}}(26)^{n}+\dots+{{n+1} \choose {2}}(26)^{2}}+26n+27 $

Can you figure out the expression in $\color{green}{green}$ is divisible by $(13)^2$

GraduateStudent
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  • okay so you solved the question without number theory by expanding, but what about $26n + 27$, they must be divisible by 13 as well right? – Patrick Schick Nov 16 '20 at 06:55
  • @Calmantec, you are trying to prove that $3^{3n+3} - 26n - 27$ must be divisible by ${13}^2$. Note that $3^{3n+3}$ is a power of the prime $3$ and hence is not divisible by $13$. – Jose Arnaldo Bebita Dris Nov 16 '20 at 06:59
  • @Calmantec, You can see that every summand in $\color{green}{green}$ is divisible by $(26)^2=4.169$. In particular the expression is divisible by 169. Thus $3^{3n+3}=169K+26n+27$. Thus $3^{3n+3}-26n-27=169K$. This proves your result? – GraduateStudent Nov 16 '20 at 07:01
  • @SunShine yeah that makes sense. Thanks a lot! – Patrick Schick Nov 16 '20 at 07:04
  • @Calmantec, I used number theory here. You can use induction here as well but that won't be fun. – GraduateStudent Nov 16 '20 at 07:05
  • @Calmantec Sunshine's last comment has serious subtlety behind it. Often, with a problem like this, I first invest 20 minutes looking for elegance. If no joy, I invest 40 minutes to an hour looking for kludgy induction. If it works, I am done, regardless of the kludginess. If no joy (again), I re-evaluate. – user2661923 Nov 16 '20 at 08:34
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Of course, SunShine's answer is great but you can do this by a simple induction without the need for the binomial theorem. The following might give you some guidance on how to think about these problems (as that was part of your original question):

The case for $n = 1$ is clear.

Then $169 | 3^{3n+3} -26n-27$ for some $n$.

Since the inductive step is to show that $P(n) \implies P(n+1)$ (this is done by replacing $n+1$ with $n$), we must show that $169 | 3^{3n+6} -26n-53$.

Since $169|3^{3n+3} -26n-27$ and we must get the right exponent for the term with base $3$, we want to multiply by $3^3$. This is allowed since it is obvious that if $a|b$, then $a|bc$.

Hence, $169|(3^{3n+3} -26n-27)3^{3}=3^{3n+6}-26\cdot27n - 27^{2}$. To get $169$ to divide our desired expression, we must figure out a way to convert one to the other using congruences. It makes sense then to check $27^2 \pmod {169}$.

You may find this is conveniently congruent to $53 \pmod {169}$ either by multiplying it out, or if you really want to use the binomial theorem the fact that $27^{2} = (2\cdot 13 +1)^{2}$ (But it is easy with exponent $2$).

So we have reduced down to the statement that $169|(3^{3n+3} -26n-27)3^{3}=3^{3n+6}-26\cdot27n - 53$.

By either doing the simple computation $26\cdot 27 \pmod {169}$ or using that $n(n+1)=n^{2}+n$,

then $26\cdot 27 = 26^{2}+27 = (13\cdot2)^{2} + 27 = 13^{2}2^{2} + 27 = 169\cdot2^{2} + 27 \equiv 27 \pmod {169}$.

Altogether, this means that $169 | 3^{3n+6} -26n-53$ and so we have proved that $P(n) \implies P(n+1)$.

Derek Luna
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A number divisible by $169=13^2$ should be divisible by $13$: and sure enough, $3^3=27\equiv1\pmod{13}$, so you get $$ 3^{3n+3}-26n-27\equiv 27^{n+1}-1\equiv 1-1\equiv0\pmod{13} $$ Of course, this doesn't help in proving the statement, but it's a step forward.

Our statement can be rewritten as $$ 169 \mid (27^{n+1}-26(n+1)-1) $$ and we can change $n+1$ into $n$: it doesn't harm, because $$ 169 \mid (27^{n}-26n-1) $$ certainly holds for $n=0$. Good, now we can go by induction. Conpute $$ (27^{n+1}-26(n+1)-1)-(27^n-26n-1)=27^{n+1}-27^n-26=26\cdot(27^n-1) $$ Now we just need to show that this number is divisible by $169$, which is equivalent to $27^n-1$ being divisible by $13$. OK, we already know this.

egreg
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