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 Sum Powers Sum Square II (Posted on 2009-07-08)
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)

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 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   nextOK`
`run 2002 1121221382103764183821126173025080062521794630945991406716447985181412364640986634625644029605963154217232387342761194114694452110217827472095636090006497251358691300247190234381726358869531688197548115137338732859195463512171761366673017832426429237997119107013304675868599898492415152827501603730179030091251649501801352593773237156327285731454441176762001979655338142876191485666369934489856131673019595693314418145771697068929726974217316242690828998377950116411669509605367891007422675363443658034720511968591074370394031117377266157313625740076978761478862585945535573933769193884396336437002240000 33484643974570853779638282783125056580043900365797925232617199636573470347654253827912449337990495566487333528673535838287098290177884813862451804920933046262295524296325721829440858140819909818368606819228270234323693566460621148622392324831490821608034988992770444273938843223914451208866267712716800.0Overflow in 25?p 8544OK`

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

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