Let us consider 3 cases: 1) x=0, This is a solution, as both L.H.S and R.H.S will be equal ( 0 ) when x=0. ( 1 solution) 2) x>0 ⇒2x(x2+1)=5x2 ⇒2(x2+1)=5x ⇒2x2−5x+2=0⇒2x2−4x−x−2=0 ⇒2x(x−2)−1(x−2)=0 ⇒(x−2)(2x−1)=0⇒x=2 or 1∕2⇒(1 integer solution ) 3) x<0 ⇒−2x(x2+1)=5x2 ⇒2x2+5x+2=0 ⇒2x2+4x+x+2=0 ⇒2x(x+2)+1(x+2)=0 ⇒(x+2)(2x+1)=0⇒x=−2 or −1∕2⇒ (1 integer solution ) So, the total number of integer solutions are 0,2,−2⇒3.