Is it valid to subtract in the middle of the proof here? Question: Prove $5^{2n+1} + 11^{2n+1} + 17^{2n+1}$ is divisible by $33$

Proposed Solution (textbook):
Let $p = 2n+1.$ Assume that $5^p + 11^p + 17^p$ divides by $33$ and prove that $5^{p+2} + 11^{p+2} + 17^{p+2}$ divides by $33$ which means we need to prove that $25\cdot5^p+121\cdot11^p+289\cdot17^p$ divides by 33.
Since we know that $5^p + 11^p + 17^p$ divides by $33,$ we can subtract it $25$ times and all we have to prove is that $96\cdot11^p+264\cdot17^p$ divides by $33.$
[Rest of proof omitted here as obvious & unrelated to my query.]

My Question:
The last procedure ("subtract it 25 times") seems odd. Perhaps it works for this particular problem, but I've never come across such methods before. Is this approach even valid?

EDIT: I now realise it's much easier to explicitly write down the factorisation step as well - i.e letting $f(k) = 25\cdot5^p+121\cdot11^p+289\cdot17^p$ ----> $f(k+1) = 25\cdot[5^p + 11^p + 17^p] - (121-25)\cdot11^p - (289-25)\cdot17^p$) ----> $f(k+1) = 25\cdot f(k) - 33\cdot[32\cdot11^{p-1} - 8\cdot17^p$]
 A: It’s fine.  $a|b\iff a|b-25ka$, because $b=ca\iff b-25ka=(c-25k)a. $
It’s easier to see with modular arithmetic though. 
Mod $3$ it’s $(-1)^{2n+1}+(-1)^{2n+1}+(-1)^{2n+1}\equiv-1-1-1=-3\equiv0,$
and mod $11$ it’s $(-6)^{2n+1}+0+6^{2n+1}\equiv6^{2n+1}-6^{2n+1}=0.$
A: That's fine.
If you eventually prove that $96*11^p+264*17^p = 33m$, and by assumption $5^p+11^p+17^p = 33n$, then this sum
$$33m + 25\times 33n$$
is still divisible by 33.
A: Hint:
The idea is to eliminate one of the terms
if $f(n)=5^{2n+1}+11^{2n+1}+17^{2n+1}$
we can use $$f(m+1)-17^2f(m)=17^{2m+1}(5^2-17^2)+11^{2m+1}(11^2-17^2)$$ is divisible by $33$
as $5^2-17^2=(5-17)(5+17)$ and $11^m\cdot(11+17)(11-17), m\ge1$ are divisible by $33$
A: You are subtly using modular congruences here.  When you render $33\mid(5^p+11^p+17^p)$, you are really rendering $5^p+11^p+17^p\equiv 0\bmod 33$.  Subtracting $25$ times that from $5^{p+2}+11^{p+2}+17^{p+2}$ means you subtracted $25×0=0$ from the residue of $5^{p+2}+11^{p+2}+17^{p+2}$ , so, if the difference is proved to have residue $0,$ then the original expression $5^{p+2}+11^{p+2}+17^{p+2}$ will also have that residue, proving it is divisible by $33$.
A: Let $S=5^{2n+1} + 11^{2n+1} + 17^{2n+1}$. Since $$5^{2n+1} +  17^{2n+1}= (5+17)(5^{2n}-...+17^{2n})= 22\cdot k$$ we see $11\mid S$. 
And since, by the binomial theorem,  $$(3k-1)^{2n+1} = 3(....)-1$$ we see that $$S = 3a-1+3b-1+3c-1 = 3(a+b+c-1)$$ so $3\mid S$
