Weak induction proof, modulus

Question

I was wondering if I could get a hint on how to solve this problem using weak induction? $$7^{4n + 2} \equiv 9 \mod 10 \quad \forall \,n \geq 0.$$

I have: $$7^{4k + 2} \cdot 7^{4} = 10x + 9.$$

I don't really know what to do from there.

Answer 1

Hints:

• induction: $\;7^{4(n+1)+2}-7^{4n+2}=7^{4n+2}\left(7^4-1\right)=7^{4n+2} \cdot 2400 \equiv 9 \cdot 0 \equiv 0 \pmod{10}$

• no induction: $\;7^2 = 49 \equiv -1 \pmod{10}$

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[ EDIT ]  The following is to (critique and) finish off the posted proof attempt.

I have: $$7^{4k + 2} \cdot 7^{4} = 10x + 9.$$

What you actually have is:

• $7^{4k + 2} = 10x + 9\;$ by the induction hypothesis

• $7^4 = 2401 = 10 y + 1\;$ by direct calculation

Then $7^{4(k+1) + 2} = (10x+9)(10y+1)= 10(10 xy + x + 9y)+9=10z+9\,$, which completes the induction step.

Answer 2

Let $$7^{4k+2}=9 \pmod{10}$$

Then for $k+1$

$$7^{4(k+1)+2}=7^{4k+2} \cdot 7^4= (10m+9) \cdot 7^4=9.7^4 \pmod{10}=(-1)(-3)^4 \pmod{10} $$

$$7^{4(k+1)+2}=-81 \pmod{10}=9 \pmod{10}$$

Therefore : $7^{4k+2}=9 \pmod{10} \implies 7^{4(k+1)+2}=9 \pmod{10}$