Question

Daca se poate aș vrea și cu explicație...​

Answer 1

 

$\displaystyle\bf\\a, b \in N\\a, b~~sunt~cifre\\a>b>0\\\\n=\Big(\overline{ab}\Big)^2-\Big(\overline{ba}\Big)^2\\\\n=\Big(10a+b\Big)^2-\Big(10b+a\Big)^2\\\\n=\Big((10a)^2+2\times10a\times b+b^2\Big)-\Big((10b)^2+2\times10b\times a+a^2\Big)\\\\n=\Big(100a^2+20a\times b+b^2\Big)-\Big(100b^2+20b\times a+a^2\Big)\\\\n=100a^2+\underline{20a\times b}+b^2-100b^2-\underline{20b\times a}-a^2\\\\n=100a^2+b^2-100b^2-a^2\\\\n=100a^2-a^2-100b^2+b^2\\\\n=99a^2-99b^2\\\\\boxed{\bf~n=99(a^2-b^2)~\vdots~99}$