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# Fast Modular Arithmetic

What is the fastest way to show that $$102^2 \equiv 30\mod {247}$$?

Note by Axas Bit
1 year, 4 months ago

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I suppose one way would be to first note that $$102 = 124 - 22$$, and so

$$A = 102^{2} = (124 - 22)^{2} = 124^{2} - 2*22*124 + 22^{2} = 62*248 - 22*248 + 484.$$

Now $$248 \equiv 1 \pmod{247}$$ and $$484 = 2*248 - 10 \equiv -10 \pmod{247}$$, so

$$A \equiv (62*1 - 22*1 - 10) \pmod{247} \equiv 30 \pmod{247}$$. · 1 year, 4 months ago

Define "fastest".

In terms of no need to think about what to do: $$102^2 = 42 \times 247 + 30$$. Staff · 1 year, 4 months ago

'Fastest way' should definitely be checking if 247 divides $$102^2-30$$. We find that it is true very fast. · 1 year, 4 months ago

Brian Charlesworth gave a standard answer. Here's an alternative approach:

Notice that $$50^2 = 2500 = 30 + 2470 = 30 + 247(10)$$, then $102^2 - 50^2 = (102 - 50)(102 + 50) = 52 \times 152 = 247 \times (\ldots ).$

This tells us that $$102^2 - 50^2 \equiv 0 \pmod {247} \Rightarrow 102^2 \equiv 50^2 \pmod{247} \Rightarrow 10^2 \equiv 30 \pmod{247}$$. · 1 year, 4 months ago

@Calvin Lin @Pi Han Goh @Otto Bretscher · 1 year, 4 months ago

@Brian Charlesworth Any thoughts? · 1 year, 4 months ago

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