IIT JEE , AIEEE Forum - Mathematics , Algebra

ananth_patri

The no. of solutions to the equⁿ x²- y² =666 where x and y are integers.....Plz answer quickly ......along with procedure....

akhil_o

The factors are 37,2,3,3

now they can be split into two in these ways:

(37,12), (111,6),(2,333),(3,222) and (74,9)

x²-y²=(x+y)(x-y)

so all factors shd be of the type (x+y)=a,(x-y)=b

adding these equations we get

x=(a+b)/2, y= (a-b)/2

But x and y have to be integers

Hence (a+b) should be even

however in the above cases, the sum is always odd

Hence x, y are never integers

Hence we have 0 solns of the equation

rate if right

nudge if wrong

akhil_o

PS i have derived x=(a+b)/2
hence for x to be an integer, (a+b) must be divisible by 2 or even
hence i said there are no real solutions

rv_hbk

SHORTCUT:

x²- y² =666 or(x+y)(x-y)=666

Product of 2 nos. is even if either both are even or one even,one odd.

For sum to be even,

either both nos. must be even of both must be odd.

If we check up,only 4*4 or 6*6 gives last digit ending in 6.

Hence,its never possible that x&y both are even as sum & diff cannot end in the same no. unless one of x or y ends in 0.

Hence no. of soln=0