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I'm starting to solve some problems of congruence and integer division, so the exercise is quite simple but I'm not sure I'm on the right track. I need to prove that the following is true for all $n \in \Bbb N$: $$9\ |\ 7 \cdot 5^{2n}+ 2^{4n+1}$$

This is what I have so far:

$$7 \cdot 5^{2n} + 2^{4n+1} \equiv 0 \ (9)$$

So I try to see what each side of the sum is congruent to: $7 \equiv -2 \ (9)$ and $5^{2n} \equiv 4^{2n} (9)$, hence: $7 \cdot 5^{2n} \equiv -2 \cdot 4^{2n} \ (9)$ and the left side is also congruent to: $-2 \cdot 4^{2n} \equiv 7^n \cdot -2 \ (9)$ which leaves me with:

$$7 \cdot 5^{2n} \equiv 7^n \cdot -2 \ (9)$$

As for the other side:

$$2^{4n+1} \equiv 7^{4n} \cdot\ 2\ (9)$$

Finally combining them:

$$7 \cdot 5^{2n} + 2^{4n+1} \equiv 7^n \cdot (-2) + 7^{4n} \cdot\ 2\ (9)$$

Am I right so far? Any hint on how to continue? Thanks!

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    $\begingroup$ Having got as far as $7\cdot5^{2n}=7\cdot4^{2n}$ and $2^{4n+1}=2\cdot4^{2n}$, can you not see a better way of continuing? $\endgroup$ – almagest May 4 '16 at 16:18
  • $\begingroup$ @almagest I couldn't see it $7 \cdot 5^{2n} + 2^{4n+1} \equiv 4^{2n} \cdot 7 + 4^{2n} \cdot\ 2\ = 4^{2n} \cdot 9\ (9)$. Thanks. $\endgroup$ – jrs May 4 '16 at 17:10
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Alternative proof by induction.


First, show that this is true for $n=0$:

$7\cdot5^{2\cdot0}+2^{4\cdot0+1}=9$

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

$7\cdot5^{2n}+2^{4n+1}=9k$

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

$7\cdot5^{2(n+1)}+2^{4(n+1)+1}=$

$16\cdot(\color\red{7\cdot5^{2n}+2^{4n+1}})+63\cdot5^{2n}=$

$16\cdot\color\red{9k}+63\cdot5^{2n}=$

$9\cdot(16k+7\cdot5^{2n})$


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

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  • $\begingroup$ thanks, it seems easier doing the proof by induction... yet I don't understand how you go from $7\cdot5^{2(n+1)}+2^{4(n+1)+1}$ to $16\cdot(\color\red{7\cdot5^{2n}+2^{4n+1}})+63\cdot5^{2n}$. I'm missing something, sorry! $\endgroup$ – jrs May 4 '16 at 17:30
  • $\begingroup$ @Gio: Yes, I just cut it shorter to make the answer a little "cleaner". All you need is a few additional computations. I'm pretty sure that you can handle it, but just in case: $7\cdot5^{2(n+1)} = 7\cdot25\cdot5^{2n} = 175\cdot5^{2n} = (16\cdot7+63)\cdot5^{2n}$, and $2^{4(n+1)+1} = 2^{4n+1+4} = 2^{4n+1}\cdot2^{4} = 2^{4}\cdot2^{4n+1} = 16\cdot2^{4n+1}$. $\endgroup$ – barak manos May 4 '16 at 17:36

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