Determinati toate numerele naturale n pentru care n^2+4n+2019 este un patrat perfect
Răspunsuri la întrebare
Răspuns:
n∈{1005;197; 69;15}
ne rezulta 1008²; 204²;84² si , rerspectiv, 48²
Explicație pas cu pas: n
n^2+4n+2019 >n^2+4n+4
n²+4n+4 =(n+2)²patrat perfect
p.p sunt printre numerele
(n+3)²=n²+6n+9=n²+4n+2019...2n=2010..n=1005
(n+5)²=n²+10n+25=n²+4n+2019.....6n=1994...n∉N
(n+6)²=n²+12n+36=n²+4n+2019////8n=1983, ..n∉N
(n+7)²=n²+14n+49=n²+4n+2019////10n=1970...n=197
(n+8)²=n²+16n+64=n²+4n+2019...12n= unmar impar..n∉N
(n+9)²=n²+18n+81==n²+4n+2019...14n=1938..n∉N
sarim (n+nr.par)²
(n+11)²=n²+22n+121==n²+4n+2019.......18n=1898..n∉N
(n+13)²=n²+26n+169==n²+4n+2019............22n=1850..n∉N
(n+15)²=n²+30n+225==n²+4n+2019.........26n=1794... .n=69
(n+17)²=n²+34n+289==n²+4n+2019........30n=1730...n∉N
(n+19)²=n²+38n+361=n²+4n+2019.......34n=1658...n∉N
(n+21)²=n²+42n+441=n²+4n+2019......38n=1378...n∉N
(n+23)²=n²+46n+529=n²+4n+2019........42n=1490...n∉N
(n+25)²=n²+50n+625=n²+4n+2019.......46n=1394...n∉N
(n+27)²=n²+54n+729=n²+4n+2019........50n=1290...n∉N
(n+29)²=n²+58n+841=n²+4n+2019..........54n=1178,,,,,n∉N
(n+31)²=n²+62n+961=n²+4n+2019........58n=1058.....n∉N
(n+33)²=n²+66n+1089=n²+4n+2019......62n=930...n=15
(n+35)²=n²+70n+1225=n²+4n+2019......66n=794....n∉N
(n+37)²=n²+74n+1369=n²+4n+2019......70n=650,,,,n∉N
(n+39)²=n²+78n+1521=n²+4n+2019.........74n=498...n∉N
(n+41)²=n²+82n+1681=n²+4n+2019.........78n=338..n∉N
(n+43)²=n²+86n+1849=n²+4n+2019........82n=170......n∉N
(n+45)²=n²+90n+2025=n²+4n+2019....96n=- 6<0, ne oprim