Tìm số tự nhiên a biết rằng a+12 và a- 77 là các số chính phương
$DK:77<a;a\in N$
Đặt $a+12=m^2;a-77=n^2$ với $m,n\in N;m>n$
ta có $m^2-n^2=a+12-(a-77)=89$
$\Leftrightarrow (m-n)(m+n)=89$ mà $m,n\in N\Rightarrow m+n;m-n\in N$ và $89=1.89$
$\Rightarrow (m-n)(m+n)=1.89$ mà $m+n>m-n$
$\Rightarrow \begin{cases}m+n=89 \\ m-n=1 \end{cases}$
$\Rightarrow \begin{cases}m=45 \\ n=44 \end{cases}$
Thay vào $a+12=m^2$
$\Rightarrow a=2013$ (nhận)
Vậy $a=2013$
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