prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. 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? 
 A: You have shown divisibility by 15.
To show divisibility by 30,
just note that the expression is even
(odd-odd+even).
A: 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.
A: Much the simplest method is to use arithmetic modulo 2,3 and 5.
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 1^{2n+1} - 1^{2n+1} - 0^{2n+1}\equiv 0 \pmod 2$$
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 2^{2n+1} - 0^{2n+1} - 2^{2n+1}\equiv 0 \pmod 3$$
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 0^{2n+1} - (-2)^{2n+1} - 2^{2n+1}\equiv 0\pmod 5$$
A: 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.
A: Use parity to deduce it is even. Or explicitly $\ \underbrace{105(\overbrace{2i\!+\!1}^{\textstyle 5^{N}}) - 15 (\overbrace{2j\!+\!1}^{\textstyle  3^{N}})}_{\textstyle 21\times 5^{N+1} - 5\times 3^{N+1}}\, =\, \overbrace{210 i - 30j  + 90}^{\textstyle \color{#c00}{30}\,(7i  -j + 3)}$
A: 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.
