20070513, 21:14  #1 
May 2004
New York City
5·7·11^{2} Posts 
Divisible by 7
Show that 2222^{5555} + 5555^{2222} is divisible by 7.
More generally, when is a^{b} + b^{a} prime? (This part I don't have a completely general answer to.) 
20070513, 22:05  #2 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
2222  2100 = 122, 122  70 = 52, 52  49 = 3, so 2222%7 = 3. 3^6 == 1 (mod 7), 5555  4800 = 755, 755  720 = 35, 35  30 = 5, so 5555%6 = 5. 2222^5555 = 3^5 == 5 (mod 7). (The mod reduction is a bit longwinded, but I wanted to do this without electronic aid)
Same for 5555^2222 == 4^2 == 2 (mod 7). 5 + 2 == 0 (mod 7) Iirc, Paul Leyland has something to say about the more general question. Alex Last fiddled with by akruppa on 20070604 at 15:51 Reason: fixed /spoiler 
20070514, 07:44  #3  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10950_{10} Posts 
Quote:
Note that site badly needs updating with more results sent to me relatively recently. Paul 

20070514, 22:05  #4 
Nov 2005
2×7×13 Posts 
Yeah, it's pretty clear that you use the properties of modulas and exponentiation together.
Last fiddled with by nibble4bits on 20070514 at 22:05 
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