Question

find all integers x and y satisfying ( x- y)^2+2y^2=27

Answer 1

We have

(x-y)² +2y² = 27

Subtracting 2y² from both sides

(x-y)² = 27 - 2y²

Now as we have (x-y)² on the left, we know a square cannot be negative.

So

(x-y)² ≥ 0

27-2y² ≥ 0

27 ≥ 2y² or 2y² ≤ 27

y² ≤ 13.5

But y is an integer

So probable values of y = ±1, ±2, ± 3 or 0

for y = 0 we get

(x-0)² = 27-0

x² = 27

But x is an integer

so x ² cannot be 27

For y = 1 we have

(x-1)² = 27 -2=25

x-1 = ±5

x= 6 or x = -4

for y = -1

(x+1)² = 27-2=25

x+1=±5

x= 4 or x = -6

for y= 2

(x-2)² = 27 -2(2)² = 19

19 is not a perfect square so y cannot be 2

In a similar way y cannot be -2

for y = 3

(x-3)² = 27 - 2(3)² = 9

x-3 = ±3

x=6 or x = 0

For y = -3

(x+3)² = 27 - 2(3)²=9

x+3 = ±3

x=0 or x = -6

The possible values are

(x,y) = {(6,1), (-4,1), (4,-1), (-6, -1), (6,3), (0,3), (0,-3),(-6, -3)}