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Sum Powers Sum Square II (Posted on 2009-07-08) Difficulty: 3 of 5
Determine all possible positive integer(s) P such that:

21994 + 21998 + 21999 + 22000 + 22002 + 2P

is a perfect square.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re: oops slight error | Comment 5 of 6 |
(In reply to oops slight error by Daniel)

computer calculation shows the only solution between p=1 and p=8543 is p=2002:

list
   10   for P=1 to 30000
   20   V=2^1994+2^1998+2^1999+2^2000+2^2002+2^P
   25   Sr=int(sqrt(V)+0.5):if Sr*Sr=V then
   30   :print P:print V
   40   :print sqrt(V)
  100   next
OK
run
 2002
 1121221382103764183821126173025080062521794630945991406716447985181412364640986
63462564402960596315421723238734276119411469445211021782747209563609000649725135
86913002471902343817263588695316881975481151373387328591954635121717613666730178
32426429237997119107013304675868599898492415152827501603730179030091251649501801
35259377323715632728573145444117676200197965533814287619148566636993448985613167
30195956933144181457716970689297269742173162426908289983779501164116695096053678
91007422675363443658034720511968591074370394031117377266157313625740076978761478
862585945535573933769193884396336437002240000
 3348464397457085377963828278312505658004390036579792523261719963657347034765425
38279124493379904955664873335286735358382870982901778848138624518049209330462622
95524296325721829440858140819909818368606819228270234323693566460621148622392324
831490821608034988992770444273938843223914451208866267712716800.0
Overflow in 25
?p
 8544
OK

  Posted by Charlie on 2009-07-10 12:29:49
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