Question

Gasiti doi divizori ai numarului 48 , a caror suma sa fie patrat perfect

Answer 1

    
[tex]D_{48} = \{1;~2;~3;~4;~6;~8;~12;~16;~24;~48\} \\ \\ 1+3 = 4 = 2^2=pp \\ 1+8=9=3^2=pp \\ 4+12 = 16 = 4^2 =pp \\ 1+24 = 25=5 ^2=pp\\ 12+24=36=6^2=pp \\ 1+48 =49=7^2 =pp \\ 16+48 = 64 = 8^2 = pp \\ [/tex]