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Given the following recurrently defined sequence of integers:

a0 = 3,  
an = 5(an−1) + 8

Prove by induction that all elements in this sequence are congruent to 3 modulo 4, or in other words: 

∀n ≥ 0 :   
an ≡ 3 (mod 4)

Basis step:

a0 = 3

5(an - 1)+ 8 ≡ 3 (mod 4)
5(a1 - 1)+ 8 ≡ 3 (mod 4)
5(a0)+ 8 ≡ 3 (mod 4)
5(3)+ 8 ≡ 3 (mod 4)
15 + 8 ≡ 3 (mod 4) 16 + 7 ≡ 3 (mod 4)
4^2 + 4 + 3 ≡ 3 (mod 4)
3 ≡ 3 (mod 4)

Induction Hypothesis:

n = k

5(a1 - 1)+ 8 ≡ 3
5(a2 - 1)+ 8 ≡ 3
.
.
.
5(ak - 1)+ 8 ≡ 3

Induction step:

n = k + 1

5((ak+1)-1)+ 8 ≡ 3 (mod 4)
5(ak) + 8 ≡ 3 (mod 4)

?????

I don't know how to proceed at this point. I know it'll be congruent if it's divisible by the modulo, which is 4 in this case. But, I don't know how to prove that with what I got so far

Kate
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3 Answers3

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I'm used to seeing induction applied in a less 'systematic' way than you did, but let me try to shed some light on it for you. Mathematical induction is like a game of dropping dominoes. If you can guarantee that

  1. The first piece will fall
  2. If one piece falls, then the next one will fall too

then all pieces will fall. The first one is the basis step, wich you've already done. The second one is the induction hypoteses and the induction step. The induction hypoteses is equivalent to "If one piece falls", you're assuming that your formula is valid for an arbitrary natural number $k$. So that's a given. The induction step is equivalent to "then the next one will fall too", so using the given fact that it works for $k$, it must work for $k+1$ too.

Now to your case. The induction hypoteses gives us that $a_k = 5a_{k-1} + 8$ is congruent to three modulo 4, so $a_k \equiv3\;(\textrm{mod}\; 4)$. Now we need to evaluate if it is true for $a_{k+1}$. We need: $$a_{k+1} \equiv3\;(\textrm{mod}\; 4)$$ But we have: $$a_{k+1} = 5a_k + 8$$ And: $$8 \equiv 0 \;(\textrm{mod}\; 4)$$ $$5 \equiv 1 \;(\textrm{mod}\; 4)$$ Then: $$a_{k+1} = 5a_k + 8 \equiv 1a_k + 0 \equiv a_k \equiv 3\;(\textrm{mod}\; 4)$$ And now we end the proof by stating: "By the Principle of Mathematical Induction, the statement is valid for all $n \geq 0$". Q.E.D

Henrique Augusto Souza
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$a_n = 3 \mod 4$ so there exists M so that $4*M + 3 = a_n$

$a_{n+1} = 5(a_{n+1}) + 8 = 5(4*M + 3) + 8 = 20M + 15 + 8 = 20M + 23 = 4(5M + 5) + 3 $ so $a_{n+1} = 3 \mod 4$

This is assuming you are not aware that if $a = c \mod b$ then $ha = hc \mod b$.

fleablood
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$$a_{k+1} - 3 = 5(a_k)+8-3$$

$$=5(a_k)+5$$

$$=5(a_k - 3) +5+15$$ $$=5(a_k - 3) +20$$ $$=5(4)(c) +20$$ for some integer c, since 4 divides $a_k -3$. So we have $$a_{k+1}-3 =4(5c+5)$$ and 4 divides $a_{k+1}$, and $a_{k+1} \equiv 3$

Kyle Funk
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