|y|+|x|=1

1) To satisfy the relation |x|+|y|=1 both x and y must be between -1 and 1.

If (a,b) satisfies this relation then so does
(b,a), (-a,b) (-b,a) (-a,-b), (-b,-a), (a,-b), and  (b,-a).
Hence this graph is the square joining the points (1,0), (0,1), (-1,0), (0,-1) and (1,0) in this order.

Alternatively the graph is the line segments (where -1 < x < 1 and -1 < y < 1) given by: x+y=1 in the first quadrant; -x+y=1 in the second quadrant; -x-y=1 in the third quadrant and x-y=1 in the fourth quadrant. (2) As 21/n must be greater than 1 we take 21/n=1+d where d > 0. Expanding by the Binomial Theorem, because all the terms are positive, we have:
2=(1+d)n > 1+nd.
Hence 1 > nd and so d < 1/n. It follows that
1 < 21/n < 1 + 1
n
 = n+1
n
and so
n
n+1
 < 1
21/n
 < 1.
(3) Fixing one value of n in xn+yn=1:
yn
= 1-xn
nyn-1 dy
dx
= -nxn-1
dy
dx
= - xn-1
yn-1
 < 0.
As x and y are positive in the first quadrant we see that the derivative is negative which shows that y decreases as x increases.

The relation is symmetric in x and y so the graph is symmetric about the line y=x and goes through the point
( 1
21/n
 , 1
21/n
 )

. In part (2) we showed that
1
21/n

is sandwiched between
n
n+1

and 1. So the graph is confined between the two squares described. (4) By a symmetry argument exactly as in part (1) the graph of |x|n+|y|n=1 is symmetrical about the lines y=x, y=-x, x=0,  y=0.

Because
n
n+1
 ® 1

as n® ¥ we see that, as n increases, the graphs in this family are confined to a narrower and narrower band rapidly becoming almost square in shape.

x^n+y^n=1 odd n
(5) For odd values of n there are no points satisfying the relation xn+yn=1 in the third quadrant where x and y, and all odd powers of these variables, are negative.

In the second quadrant if x is large and negative then xn is large and negative and yn=1-xn is large and positive and y® -x as n® ¥.

By a similar argument, or explaining it by the symmetry of the relation, y® -x as n® ¥ in the fourth quadrant.