Question

Scrieti numerele 19,23,27 ca diferenta de numere naturale patrate perfecte

Answer 1

Orice numar impar se poate scrie in mod unic ca diferenta de doua patrate perfecte consecutive.

Vom folosi suma Gaus a sirului numerelor impare
1+3+5+7+...+k 
care este generator de patrate perfecte.

Pregatim formulele:
Numarul de termeni (n) ai sirului este:

[tex]\displaystyle\\ n = \frac{k-1}{2}+1[/tex]

[tex]\displaystyle\\ 1+3+5+7+...+k = \frac{n(k+1)}{2} [/tex]

Rezolvare:

[tex]\displaystyle\\ 19 = (1+3+5+\cdots+19) - (1+3+5+\cdots+17) \\\\ n_{19} = \frac{19-1}{2}+1=10 ~~~\text{ si }~~~n_{17}=\frac{17-1}{2}+1=9 \\\\ 19= \frac{10(19+1)}{2}- \frac{9(17+1))}{2}\\\\ 19 =\frac{10(20)}{2}- \frac{9(18)}{2}\\\\ \boxed{\bf 19 = 10^2 - 9^2=100-81}[/tex]


[tex]\displaystyle\\ 23 = (1+3+5+\cdots+23) - (1+3+5+\cdots+21) \\\\ n_{23} = \frac{23-1}{2}+1=12 ~~~\text{ si }~~~n_{21}=\frac{21-1}{2}+1=11 \\\\ 23= \frac{12(23+1)}{2}- \frac{11(21+1))}{2}\\\\ 23 =\frac{12(24)}{2}- \frac{11(22)}{2}\\\\ \boxed{\bf 23 = 12^2 - 11^2=144-121}[/tex]


[tex]\displaystyle\\ 27 = (1+3+5+\cdots+27) - (1+3+5+\cdots+25) \\\\ n_{27} = \frac{27-1}{2}+1=14 ~~~\text{ si }~~~n_{25}=\frac{25-1}{2}+1=13 \\\\ 27= \frac{14(27+1)}{2}- \frac{13(25+1))}{2}\\\\ 27 =\frac{14(28)}{2}- \frac{13(26)}{2}\\\\ \boxed{\bf 27 = 14^2 - 13^2=196-169}[/tex]