prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.

Question

Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.

I have tried induction as follows.

Step 1:

Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.

Try n = 1, we get: $5^{3} - 3^{3} - 2^{3} = 90$, which is also divisible by 30.

Step 2:

Assume it is true for n = k.

So we are assuming the following equality is true: $5^{2k+1} - 3^{2k+1} - 2^{2k+1} = 30M$, for some integer M.

Step 3:

Now we look at the next case: n = k + 1.

$5^{2(k+1)+1} - 3^{2(k+1)+1} - 2^{2(k+1)+1}$

= $5^{2k+3} - 3^{2k+3} - 2^{2k+3}$

= $25\times5^{2k+1} - 9\times3^{2k+1} - 4\times2^{2k+1}$

= $21\times5^{2k+1} + 4\times5^{2k+1} - 5\times3^{2k+1} - 4\times3^{2k+1} - 4\times2^{2k+1}$

= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times[5^{2k+1} - 3^{2k+1} - 2^{2k+1}]$

= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times30M$. (Assumed in step 2)

The last term is divisible by 30. But I cannot get a factor 30 out of the first two terms. I can show divisibility by 15 as follows:

= $7\times3\times5\times5^{2k} - 5\times3\times3^{2k} + 4\times30M$

= $7\times15\times5^{2k} - 15\times3^{2k} + 4\times15\times2M$

But how do I show divisibility by 30?

Answer 1

You have shown divisibility by 15.

To show divisibility by 30, just note that the expression is even (odd-odd+even).

Answer 2

Let $a_n = 5^{2n+1} - 3^{2n+1} - 2^{2n+1} = 5 \cdot 25^n - 3 \cdot 9^n - 2 \cdot 4^n$. Then $a_{n+3} = 38 a_{n+2} - 361 a_{n+1} + 900 a_n$ (*). Therefore, you only need to check the claim for $n=0,1,2$, which is immediate.

(*) Because $(x-25)(x-9)(x-4) = x^3 - 38 x^2 + 361 x - 900$. The coefficients are not important here. The important point is that $a_n$ satisfies a linear recurrence with integer coefficients.

Answer 3

Much the simplest method is to use arithmetic modulo 2,3 and 5. $$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 1^{2n+1} - 1^{2n+1} - 0^{2n+1}\equiv 0 \pmod 2$$ $$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 2^{2n+1} - 0^{2n+1} - 2^{2n+1}\equiv 0 \pmod 3$$ $$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 0^{2n+1} - (-2)^{2n+1} - 2^{2n+1}\equiv 0\pmod 5$$

Answer 4

You could show by induction that modulo $30$

$5^{2k+1}\equiv5,$

$3^{2k+1}\equiv3$ if $k$ is even and $-3$ if $k$ is odd, and

$2^{2k+1}\equiv2$ if $k$ is even and $8$ if $k$ is odd.

Answer 5

Use parity to deduce it is even. Or explicitly $\ \underbrace{105(\overbrace{2i\!+\!1}^{\textstyle 5^{N}}) - 15 (\overbrace{2j\!+\!1}^{\textstyle 3^{N}})}_{\textstyle 21\times 5^{N+1} - 5\times 3^{N+1}}\, =\, \overbrace{210 i - 30j + 90}^{\textstyle \color{#c00}{30}\,(7i -j + 3)}$

Answer 6

You can show divisiblity by $30$ by showing divisibilty by $2,3$ and $5$

So show $21\times5^{2k+3} - 5\times3^{2k+3} + 4\times30M$ is divisible by $2,3$ and by $5$

So

$21\times5^{2k+3} - 5\times3^{2k+3} + 4\times30M=$

$15(7\times 5^{2k+2} - 3^{2k+2} + 4\times 2M)$.

Obviously $15$ is divisible by $3$ and by $5$.

So just remains to show $ 7\times 5^{2k+2} - 3^{2k+2} + 4\times 2M$ is divisible by $2$ (or in other words even).

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But it's easier without induction.

$5|5^{2n+1}$ and so need to show $5|3^{2n+1}+ 2^{3n+1}$.

This is the "freshman dream". but if you know modular arithmetic.

$3^{2n+1} \equiv (-2)^{2n+1} \equiv -2^{2n+1} \pmod 5$ so $3^{2n+1}+2^{2n+1} \equiv 0 \pmod 5$.

But if you dont know modular arithmetic: $3^{2n+1}+ 2^{2n+1} = (3+2)(3^{2n}- 2*3^{2n-1} + ...... - 2^{2n-1}*3 +2^{2n})$ and $3+2 = 5$.

...

Likewise $3|3^{2n+1}$ so we need to show $3|5^{2n+1}-2^{2n+1}$.

That the same thing $5 \equiv 2 \pmod 3$ so $5^{2n+1}\equiv 2^{2n+1}\pmod 3$ so $3|5^{2n+1} -2^{2n+1}$.

Or $5^{2n+1} -2^{2n+1}= (5-2)(5^{2n} + 5^{2n-1}*2 + .... + 5*2^{2n-1}+ 2^{2n})$. And $5-2=3$.

....

And $2|2^{2n+1}$ so we have to show $2|5^{2n+1} - 3^{2n+1}$ which we can do the exact same ways as above, or we can not that ODD minus ODD is even.

So $2,3,5$ each divide $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ so $30$ does as well.