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Find four prime divisors < 100 for 332−232.
Source: British Math Olympiad, 2006.
This factors nicely as 332−232=(316+216)(316−216), and we can continue factoring in this way to get 332−232=(316+216)(38+28)(34+24)(32+22)(32−22).The final three terms are 5,13,97, so we have three of the four required primes. For another prime divisor, consider 316−216. By Fermat's Little Theorem a16−1≡0mod17 for all a with (a,17)=1, and so it follows that 316−216≡0mod17, and we therefore have 17 as a fourth such prime divisor.
Alternatively, note (317)=−1,(217)=1, hence by Euler's Criterion 38≡−1mod17 and 28≡1mod17, giving 38+28≡0mod17.
Source: British Math Olympiad, 2006.
This factors nicely as 332−232=(316+216)(316−216), and we can continue factoring in this way to get 332−232=(316+216)(38+28)(34+24)(32+22)(32−22).The final three terms are 5,13,97, so we have three of the four required primes. For another prime divisor, consider 316−216. By Fermat's Little Theorem a16−1≡0mod17 for all a with (a,17)=1, and so it follows that 316−216≡0mod17, and we therefore have 17 as a fourth such prime divisor.
Alternatively, note (317)=−1,(217)=1, hence by Euler's Criterion 38≡−1mod17 and 28≡1mod17, giving 38+28≡0mod17.
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