Find the consecutive even integers whose squares have the sum 340.

Solution:

Let two consecutive even integer be $2 x$ and other $(2 x+2)$

Then according to question

$(2 x)^{2}+(2 x+2)^{2}=340$

$4 x^{2}+4 x^{2}+8 x+4=340$

$8 x^{2}+8 x=340-4$

$8 x^{2}+8 x-336=0$

$8 x^{2}+8 x-336=0$

$8\left(x^{2}+x-42\right)=0$

$\left(x^{2}+x-42\right)=0$

$x^{2}+7 x-6 x-42=0$

$x(x+7)-6(x+7)=0$

$(x+7)(x-6)=0$

$(x+7)=0$

$x=-7$

or

$(x-6)=0$

$x=6$

Since, x being a positive number, so x cannot be negative.

Therefore,

When $x=6$ then even integer

$2 x=2 \times 6$

$=12$

And

$2 x+2=2 \times 6+2$

$=14$

Thus, two consecutive odd positive integer be 12,14