The number of pairs of consecutive positive even integers such that the sum of their squares is 290 is
The number of pairs of consecutive positive even integers such that the sum of their squares is 290 is
$0$
$1$
$2$
$3$
Solution
Let $x$ and $x+2$ be two consecutive positive even integers.
Given, $x^2+(x+2)^2=290$
$\Rightarrow \quad x^2+x^2+4 x+4-290=0$
$\begin{array}{ll}\Rightarrow & 2 x^2+4 x-286=0 \\ \Rightarrow & x^2+2 x-143=0\end{array}$
$\Rightarrow \quad x=-13,11$
[ -13 not postive integer and $x=13$, is not even positive integer]
Hence, number of solutions $=0$