Suppose 2
2022 - 31*2
2014 + 2
n is a perfect square for a certain positive integer value of n.
Find the value of n.
clearvars,clc
s=sym(2)^2022-31*sym(2)^2014;
for n=1:10000
sq=s+sym(2)^n;
sr=round(sqrt(sq));
if sr^2==sq
disp(n)
disp(sq)
disp(sr)
disp(' ')
end
end
finds
The answer is n = 2020.
The perfect square is
54363712869869085901338836195794959299780577837651216455768268954443194293026000
19223750107880985591913363599862454300533165545548831506280973830749096554833151
35559483314158308541352736217054705139055003702845845371074461594521389969811382
69298321485973985698298156246278140651151584570107414872308819795818028752398759
53970231324034578376789585125442948953175675406723485017759779436182457386011570
77773633284826150317413079106826537929892986219648074488179991593621631786310290
95329404822545377261021274018600872867410235882332513671173040561687006541702268
3018419204940771596566792393054977754226518654976
Its square root is
23316027292373176903837729067545639397816168702712411297976008450938838872478610
02545199672302954187285646008268595647491607228141666475358870244080254409877336
16194780175266242762583406157196047264283003650291295642743041978597182087442236
26663688902106923535445915756829095313676110665757759533718962176
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Posted by Charlie
on 2024-02-26 10:11:54 |
computer solution
