Weekly Challenge 19: Prime APs

7054 /www/nrich/html/content/id/7054/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/6206&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/6413&part=">Try this next</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/150476.html">Ask NRICH</a> </li> <li> <a href="http://en.wikipedia.org/wiki/Primes_in_arithmetic_progression"> Read all about it</a> </li> <li> <a href="http://nrich.maths.org/7036&part=solution">Last week's solution</a> </li> </ul> <div> Show that if three prime numbers, all greater than 3, form an arithmetic progression then the common difference of the progression is divisible by 6. Find some examples of three primes which include the number 3 and form an AP, and show that in every such case the common difference is not divisible by 6. </div> <div class="framework"> Did you know ... ? Explorations of arithmetic progressions of prime numbers are an ongoing subject of mathematics research.</div> <div> </div> <div>  </div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This challenge was previously published on the site as a monthly problem. A solution was sent in by Yatir of Maccabim-Reut High-School, Israel. Let's say that the 3 primes in the AP are: P, Q,and S the common difference is d. So we have: P, P+d, P+2d . We are working with primes greater than 3 so they all have to be odd and d must be even. This is because the difference between 2 odds is always even as (2n + 1) - (2k + 1) = 2(n - k) . I'm going to work modulus 6: even number have residues of: 0, 2, 4 (mod 6) and odd numbers have residues of: 1, 3, 5 (mod 6). Our prime numbers must be be congruent to 1 or 5 (mod 6), because if they were congruent to 3 they would be divisible by 3 and thus not prime numbers. Lets say that is congruent to 1 (mod 6) so is congruent to either: 1 + 0 = 1 (mod 6) 1 + 2 = 3 (mod 6) 1 + 4 = 5 (mod 6). Because P + d is a prime number it can't be congruent to 3 (mod 6) so d must be congruent to either 0 or 4 (mod 6). And is congruent to either: 1 + 2 $\times$ 0 = 1 (mod 6) 1 + 2 $\times$ 4 = 1 + 7 = 9 = 3 (mod 6) Because P + 2d is a prime number as well it can't be congruent to 3 (mod 6), so must be congruent to 0 (mod 6) Lets say that P is congruent to 5 (mod 6). So P + d is congruent to either: 5 + 0 = 5 (mod 6) 5 + 2 = 7 = 1 (mod 6) 5 + 4 = 9 = 3 (mod 6) Because P + d is a prime number it can't be congruent to 3 (mod 6) so d must be congruent to either 0 or 2 (mod 6). And is congruent to either: 5 + 2 $\times$ 0 = 5 (mod 6) 5 + 2 $\times$ 2 = 1 + 4 = 9 = 3 (mod 6) Because P + 2d is a prime number as well it can't be congruent to 3 (mod 6), so d must be congruent to 0 (mod 6) Following from all of this d must be congruent, in all cases, to 0 (mod 6), meaning it gives a remainder 0 when divided by 6. So d is divisible by 6, hence proved. Examples from APs where one of the prime numbers is 3 3, 5, 7 (d = 2) 3, 7, 11 (d = 4) 3, 11, 19 (d = 8) 3, 13, 23 (d = 10) 3, 17, 31 (d = 14) 3, 23, 43 (d = 20) In these examples none of the differences is divisible by 6 but is this true in general for AP's containing 3. Yes because if the first number is 3, and the common difference is divisible by 6, call this difference 6k, then the second number is 3 + 6k which is divisible by 3 so it is not a prime. Hence no AP of 3 primes exists which has common difference divisible by 6. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This challenge was previously published on the site as a monthly problem. A solution was sent in by Yatir of Maccabim-Reut High-School, Israel. Let's say that the 3 primes in the AP are: P, Q,and S the common difference is d. So we have: P, P+d, P+2d . We are working with primes greater than 3 so they all have to be odd and d must be even. This is because the difference between 2 odds is always even as (2n + 1) - (2k + 1) = 2(n - k) . I'm going to work modulus 6: even number have residues of: 0, 2, 4 (mod 6) and odd numbers have residues of: 1, 3, 5 (mod 6). Our prime numbers must be be congruent to 1 or 5 (mod 6), because if they were congruent to 3 they would be divisible by 3 and thus not prime numbers. Lets say that is congruent to 1 (mod 6) so is congruent to either: 1 + 0 = 1 (mod 6) 1 + 2 = 3 (mod 6) 1 + 4 = 5 (mod 6). Because P + d is a prime number it can't be congruent to 3 (mod 6) so d must be congruent to either 0 or 4 (mod 6). And is congruent to either: 1 + 2 $\times$ 0 = 1 (mod 6) 1 + 2 $\times$ 4 = 1 + 7 = 9 = 3 (mod 6) Because P + 2d is a prime number as well it can't be congruent to 3 (mod 6), so must be congruent to 0 (mod 6) Lets say that P is congruent to 5 (mod 6). So P + d is congruent to either: 5 + 0 = 5 (mod 6) 5 + 2 = 7 = 1 (mod 6) 5 + 4 = 9 = 3 (mod 6) Because P + d is a prime number it can't be congruent to 3 (mod 6) so d must be congruent to either 0 or 2 (mod 6). And is congruent to either: 5 + 2 $\times$ 0 = 5 (mod 6) 5 + 2 $\times$ 2 = 1 + 4 = 9 = 3 (mod 6) Because P + 2d is a prime number as well it can't be congruent to 3 (mod 6), so d must be congruent to 0 (mod 6) Following from all of this d must be congruent, in all cases, to 0 (mod 6), meaning it gives a remainder 0 when divided by 6. So d is divisible by 6, hence proved. Examples from APs where one of the prime numbers is 3 3, 5, 7 (d = 2) 3, 7, 11 (d = 4) 3, 11, 19 (d = 8) 3, 13, 23 (d = 10) 3, 17, 31 (d = 14) 3, 23, 43 (d = 20) In these examples none of the differences is divisible by 6 but is this true in general for AP's containing 3. Yes because if the first number is 3, and the common difference is divisible by 6, call this difference 6k, then the second number is 3 + 6k which is divisible by 3 so it is not a prime. Hence no AP of 3 primes exists which has common difference divisible by 6. </mdoxml> 2 3 0 0 0 1 1 Weekly Challenge 19: Prime APs A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Collections Weekly Challenge Admin Stage 5 - Core Mapping Stage 5 Core Mapping Document Sequences and Series AS Secondary Mapping Document DisplayCabinet