Let x and x+2 be two consecutive positive even integers. Given,x2+(x+2)2=290 ⇒ x2+x2+4x+4−290=0 ⇒ 2x2+4x−286=0 ⇒ x2+2x−143=0 ⇒ x=−13,11 [−13 not positive integer and x=13 ,is not even positive integer ] Hence, number of solutions =0