判断同余式$$5^x + 3 \equiv 5 \mod 100$$是否存在自然数解

Let's work through this problem step by step to find out.

First, let's rearrange the congruence to isolate the exponential term:
$$5^x \equiv 5 - 3 \pmod{100}$$
That simplifies to:
$$5^x \equiv 2 \pmod{100}$$

Now, let's break down the behavior of (5^x) modulo 100 for natural numbers (x) (we'll cover both (x=0) and (x \geq 1) since definitions of natural numbers can vary slightly):

• When (x=0): (5^0 = 1). (1 \mod 100 = 1), which doesn't equal 2.
• When (x=1): (5^1 = 5). (5 \mod 100 = 5), still not a match for 2.
• For (x \geq 2): Notice that (5^2 = 25), and multiplying any multiple of 25 by 5 keeps it a multiple of 25 (e.g., (25 \times 5 = 125 \equiv 25 \pmod{100}), (25 \times 5^2 = 625 \equiv 25 \pmod{100})). So for all (x \geq 2), (5^x \equiv 25 \pmod{100})—which is clearly not congruent to 2 modulo 100.

A quick sanity check: Any power of 5 is a multiple of 5, so (5^x \mod 100) must end in 0 or 5. Since 2 ends in 2, there's no way these can ever be congruent.

Conclusion: There are no natural number solutions to the congruence (5^x + 3 \equiv 5 \pmod{100}).

内容的提问来源于stack exchange，提问作者Zhenqing Xu