Solving a 'Harvard' University entrance exam question

[00:00:00]Hello Friends find the value of x & y If x^2-y^2=65 where x,y belongs to positive integers let's have a solution this problem is same as (x+y)(x-y)=65 we have choices here 65x1 or 1x65 or 13x5 or 5x13 so we have four choices by comparison of (x+y) and (x-y) (x+y) is greater than (x-y) according to that 65 is greater than 1 this is possible 1 is less than 65 not possible 13 is greater than 5 possible 5 is less than 13, not possible so possible cases here case I is 65x1 and case II is 13x5 first of all, take case I, which is 65x1 we have (x+y)(x-y)=65x1 by comparing x+y=65 and x-y=1 add them 'y' cancels 2x=66 to find 'x', divide by '2' on both sides 2x/2=66/2 2 cancels 2x33=66 x=33 put this into x-y=1 to find 'y' 33-y=1 33-1=y y=32 so (x,y)=(33,32) now, take case II which is 13x5 (x+y)(x-y)=13x5 by comparing x+y=13 and x-y=5 add them y cancels 2x=18 to find 'x', divide by '2' on both sides 2x/2=18/2, where 2 cancels 2x9=18 x=9 to find 'y', take x-y=5 9-y=5 9-5=y y=4 (x,y)=(9,4) so finally, (x,y)=(33,32), (9,4) in the next step, I'm going to verify x^2-y^2=65 put values of x and y (33)^2-(32)^2=65 and (9)^2-(4)^2=65 (33)^=33x33=1089 (32)^2=32x32=1024 1089-1024=65 65=65 L.H.S=R.H.S now, take 81-16=65 65=65 L.H.S=R.H.S which shows that (x,y)=(33,32), (9,4) satisfies this equation of x^2-y^2=65 thanks for watching this video please subscribe this channel to get the notification of my new videos and don't forget to share these videos with your classmates and friends so that they also have a benefit of it ok bye