×

# Other Approach.?

Hello friends, I solved this question but I want to know other method which u think will be the shortest method. Find the number of positive integral values of $$x \leq 100$$ such that $$3^{x} - x^{2}$$ is divisible by 5.I think answer is 20.

Note by Kiran Patel
4 years, 3 months ago

Sort by:

We are looking for the values of $$x$$ such that $$3^x-x^2=5k$$ with $$k\in\mathbb{Z}$$, or $$3^x \equiv x^2 \pmod{5}$$. The sequence $$a_x = 3^x \pmod{5}$$ for $$1 \leq x \leq 100$$ is $$3,4,2,1,3,4,2,1,\dots$$ with a period of 4. The sequence $$b_x = x^2 \pmod{5}$$ for $$1 \leq x \leq 100$$ is $$1,4,4,1,0,1,4,4,1,0,\dots$$ with a period of 5. So, if $$3^x \equiv x^2 \pmod{5} \implies a_x=b_x$$, then either $$a_x=b_x=1$$ or $$a_x=b_x=4$$.

Case 1: $$a_x=b_x=1$$. $$a_x = 1$$ whenever $$x \equiv 0 \pmod{4}$$ and $$b_x = 1$$ whenever $$x \equiv 1 \pmod{5}$$ or $$x \equiv 4 \pmod{5}$$.

$$\bullet$$ Case 1a) $$x \equiv 0 \pmod{4}$$ and $$x \equiv 1 \pmod{5} \implies x \equiv 16 \pmod{20} \implies x \in \{16,36,56,76,96\}$$.

$$\bullet$$ Case 1b) $$x \equiv 0 \pmod{4}$$ and $$x \equiv 4 \pmod{5} \implies x \equiv 4 \pmod{20} \implies x \in \{4,24,44,64,84\}$$.

Case 2: $$a_x=b_x=4$$. $$a_x = 4$$ whenever $$x \equiv 2 \pmod{4}$$ and $$b_x = 4$$ whenever $$x \equiv 2 \pmod{5}$$ or $$x \equiv 3 \pmod{5}$$.

$$\bullet$$ Case 2a) $$x \equiv 2 \pmod{4}$$ and $$x \equiv 2 \pmod{5} \implies x \equiv 2 \pmod{20} \implies x \in \{2,22,42,62,82\}$$.

$$\bullet$$ Case 2b) $$x \equiv 2 \pmod{4}$$ and $$x \equiv 3 \pmod{5} \implies x \equiv 18 \pmod{20} \implies x \in \{18,38,58,78,98\}$$.

So, $$x \in \{ 2,4,16,18,22,24,36,38,42,44,56,58,62,64,76,78,82,84,96,98 \}$$, which are indeed 20 values.

- 4 years, 3 months ago

I did the same but I want any other short method.Thanks for ur efforts......

- 4 years, 3 months ago