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.

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.

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TopNewestWe 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.

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

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