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How can I prove that $$5^{2n+1}+3^{n+2} \cdot 2^{n-1} $$ can be divided by 19 for any nonnegative n? What modulo should I choose?

  • 12
    You should try to take mod 19 :-) – 2017-01-15
  • 0
    Did you try to prove it by induction ? – 2017-01-15
  • 0
    yes indeed. 19 is prime you see – 2017-01-15
  • 7
    It's not true for $n=0$ so you had better change the hypothesis – 2017-01-15
  • 3
    If you are studying number theory (per the tags), then I highly recommend that you resist the temptation to settle for plug-and-chug brute force inductive proofs devoid of arithmetical insight. The arithmetical essence of dvisibilities like this becomes much clearer when one uses modular arithmetic. For more complex problems the brute force methods will fail miserably, whereas the modular methods remain just as simple. – 2017-01-15
  • 0
    There is a [meta question](http://meta.math.stackexchange.com/q/25735/242) about this question. Be aware that this question was on the Hot List. – 2017-01-23

6 Answers 6

2

$2^{n-1} = \frac12 2^n \equiv 10 \cdot 2^n \pmod {19}$. Hence:

$5^{2n+1} +3^{n+2} \cdot 2^{n-1} \equiv 5 \cdot 6^n + 90 \cdot 6^n \pmod {19} \equiv 95 \cdot 6^n \pmod {19} \equiv 0 \pmod {19}$

  • 0
    You should probably also observe that $5^2\equiv 6 \bmod 19$ in your preamble, then it flows nicely. – 2017-01-15
30

For $n=0$, the formula says that $\left.19\middle|\frac{19}2\right.$, which is false. So consider $n\ge1$: $$ \begin{align} 5^{2n+1}+3^{n+2}2^{n-1} &=125\cdot25^{n-1}+27\cdot6^{n-1}\\ &\equiv11\cdot6^{n-1}+8\cdot6^{n-1}&\pmod{19}\\ &=19\cdot6^{n-1}\\ &\equiv0&\pmod{19} \end{align} $$ Since $$ \begin{align} 125&\equiv11&\pmod{19}\\ 25&\equiv6&\pmod{19}\\ 27&\equiv8&\pmod{19} \end{align} $$

  • 1
    would either downvoter care to comment? – 2017-01-23
  • 2
    I don't know why you got downvoted (although the [meta discussion](http://meta.math.stackexchange.com/q/25735) *might* have something to do with it), but I did notice that, as pointed out by David Quinn above, your congruences are only valid if $n \ge 1$, since otherwise $6^{n-1}$ is not an integer. That's something that should probably be explicitly noted, even at the risk of spoiling the elegant simplicity of the answer. – 2017-02-02
  • 0
    @IlmariKaronen: If we consider $n=0$, the original formula claims that $\left.19\middle|\frac{19}2\right.$. That's like saying $2\mid1$. So the original formula does not hold for $n=0$. – 2017-02-02
17

$5^{2n+1} = 5 \times 25^n \equiv 5 \times 6^n$ (modulo 19). Hence we have $5^{2n+1} + 3^{n+2}\times2^{n-1} \equiv 6^{n-1} \times (30 + 27) = 6^{n-1}\times3\times 19$ (modulo 19)

5

Its arithmetical essence is clarified using Congruence Product and Power Rules, namely

$$\begin{align}{\rm mod}\,\ 19\!:\qquad \big[\,\ 5^{\large\color{#c00}1} \ &\equiv\, -3^{\large\color{#c00} 2}\ \cdot\, 2^{\large\color{#c00}{-1}}\big]\ \ \ {\rm by}\ \ \ \ \,2\cdot 5\equiv -3^{\large 2}\\ \times\ \ \, \big[\, 5^{\large 2n} &\equiv\ \ \ 3^{\large n}\,\cdot\, 2^{\large n}\ \ \ \big]\ \ \ {\rm by} \ \ \Big[\,5^{\large 2}\equiv 3\cdot 2\,\Big]^{\large n}\\[.1em] \Rightarrow\,\ 5^{\large 2n+\color{#c00}1}\!&\equiv -3^{\large n+\color{#c00}2} 2^{\large n\color{#c00}{-1}} \end{align}\qquad\qquad\quad\ $$

Therefore we infer that $\ 19\mid 5^{\large 2n+1}+\ 3^{\large n+2} 2^{\large n-1}$ when it is integral, i.e. for all $\,n\ge 1$

Remark $ $ It is even clearer when written in fractional form, namely

$$ \dfrac{5^{\large 2n+1}}{3^{\large n+2}2^{\large n-1} }\ \equiv\ \left[\dfrac{2\cdot 5}{3\cdot 3}\right]\left[\dfrac{5^{\large 2}}{3\cdot 2}\right]^{\large n}\!\!\equiv\, -1\cdot 1^{\large n}\equiv\, -1$$

Notice how use of congruence language greatly simplifies the inductive step, reducing it to the trivial induction that $\,1^n\equiv 1.$

4

You can prove this by induction.

First, show that this is true for $n=1$:

$5^{2\cdot1+1}+3^{1+2}\cdot2^{1-1}=19\cdot8$

Second, assume that this is true for $n$:

$5^{2n+1}+3^{n+2}\cdot2^{n-1}=19k$

Third, prove that this is true for $n+1$:

$5^{2(n+1)+1}+3^{n+1+2}\cdot2^{n+1-1}=$

$5^{2+2n+1}+3^{1+n+2}\cdot2^{1+n-1}=$

$5^{2}\cdot5^{2n+1}+3^{1}\cdot3^{n+2}\cdot2^{1}\cdot2^{n-1}=$

$5^{2}\cdot5^{2n+1}+3^{1}\cdot2^{1}\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot5^{2n+1}+\color\green{6}\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot5^{2n+1}+(\color\green{25-19})\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot5^{2n+1}+25\cdot3^{n+2}\cdot2^{n-1}-19\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot(\color\red{5^{2n+1}+3^{n+2}\cdot2^{n-1}})-19\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot\color\red{19k}-19\cdot3^{n+2}\cdot2^{n-1}=$

$19\cdot25k-19\cdot3^{n+2}\cdot2^{n-1}=$

$19\cdot(25k-3^{n+2}\cdot2^{n-1})$

Please note that the assumption is used only in the part marked red.

  • 11
    I think that such a long-winded proof is counter-productive. It hides the essentials in a mass of details. – 2017-01-15
  • 1
    @TonyK: That's what the red and green parts are for. – 2017-01-15
  • 6
    Isn't it customary to have running equals at the *start* of each line? My eyes want to read the lead coefficients as line-numbers. – 2017-01-16
  • 4
    Yes, barak -- but what are all those black parts for? – 2017-01-16
  • 4
    @TonyK To prove what needed to be proved, that's what for. You aren't suggesting that "$6=25-19$ therefore the result follows by induction" would be a sufficient proof, are you? I myself would be content with *slightly* less working than Barak provided, but there's a danger of the writer being so concise that readers are forced to do the algebra to verify the proof -- so why didn't the writer supply that algebra in the first place? – 2017-01-16
2

Denote $\mathcal{P}(n)$ the statement that $5^{2n+1} + 3^{n+2}\cdot 2^{n-1}$ is divisible by $19$. You can check for yourself that $\mathcal{P}(1)$ is true. A setup for the proof of $\mathcal{P}(n+1)$:

\begin{align} 5^{2(n+1)+1} + 3^{(n+1)+2}\cdot 2^{(n+1)-1} & = \\ 25 \times 5^{2n+1} + 6 \times 3^{n+2}\cdot 2^{n-1} & = \\ 19 \times 5^{2n+1} + 6 \times 5^{2n+1} + 6 \times 3^{n+2}\cdot 2^{n-1} & = \cdots \end{align} can you finish the proof from here?