How do I prove that $ 10 | 4^{n + 2 }+ 5^{n + 2 }+ 4^{n + 3 }+ 5^{n + 3 }$

Question

I have no idea how to prove this. I don't even have an idea where to start from, could someone drop a few hints?

You have been around for almost two months. Haven't you yet noticed that you are supposed to use MathJax around here? — José Carlos Santos, Dec 10 '20 at 10:46
The expression is not clear (I assume the exponents are $n+2$ and $n+3$). Additionally, the concrete question should be in the body, the title should just be a description what it is about. — Peter, Dec 10 '20 at 10:47
Nevertheless, a small hint : Show that the expression is both divisible by $2$ and by $5$. — Peter, Dec 10 '20 at 10:49
@JoséCarlosSantos I'm struggling with formatting as of now, I will practise later. So sorry for the inconvenience. — Ognjen Ognjanovic, Dec 10 '20 at 10:52
With a little modulo-arithmetic, you can show this without induction. — Peter, Dec 10 '20 at 10:53
@Peter Thanks, I've just simplified it to be both divisible by 2 and by 5. Thanks a lot :) — Ognjen Ognjanovic, Dec 10 '20 at 10:53

score 1 · Answer 1 · answered Dec 10 '20 at 10:53

1

Note that $ 4^{n + 2 }+ 5^{n + 2 }+ 4^{n + 3 }+ 5^{n + 3 } = 5\cdot 4^{n + 2 }+ 6 \cdot 5^{n + 2 } = 80 \cdot 4^n + 150 \cdot 5^n $.

answered Dec 10 '20 at 10:53

lhf

sirous · Accepted Answer · 2020-12-10T15:47:52.990

1

If you mean

$N=5^{n+3}+5^{n+2}+4^{n+3}+4^{n+2}$

Then we have:

$N=5^{n+2}(5+1)+4^{n+2}(4+1)= 5k$

But N is even therefore it must be a multiple of $2\times 5 =10$.

edited Dec 10 '20 at 15:47

answered Dec 10 '20 at 10:55

sirous

score 1 · Answer 3 · answered Dec 10 '20 at 12:05

1

Hint: $\ ({a,b\!+\!1})(\color{#c00}{a\!+\!1,b})\mid a^j(\color{#c00}{a\!+\!1}) + \color{#c00}{b^k}(b\!+\!1)\,$ if $\,j,\color{#c00}{k>0}$.

OP is case $\ a,b = 4,5,\ j,k = n\!+\!2$

answered Dec 10 '20 at 12:05

Bill Dubuque

Note the factors are coprime $,(d\mid (a,b!+!1),(a!+!1,b)\Rightarrow,d\mid a,a!+!1)$ so their lcm = product. – Bill Dubuque Dec 10 '20 at 12:11

3 Answers3