Show algebraically that for positive integers pp and qq, if
a = p2 − q2
b = 2p
c = p2 + q2
then a2 + b2 = c2.

a² + b² = c²

⇒ p⁴ + q⁴ + 2p²q² = p⁴ + q⁴ + 2p²q²

Step-by-step explanation:

Data provided:

a = p² − q²

b = 2p q                   (for the required result the value of b should be '2pq' instead of '2p' )

c = p² + q²

To show:

a² + b² = c²

Now,

a² = (p² − q² )²

we know,

(x + y) = x² + y² + 2xy

(x - y) = x² + y² - 2xy

Thus,

a² = (p² − q² )²

=  (p²)² + (q²)² - 2p²y²

= p⁴ + q⁴ - 2p²q²       (1)

and,

b² = (2pq)² = 4p²q²         (2)

also,

c² =   (p² + q² )²

= (p²)² + (q²)² + 2p²y²

= p⁴ + q⁴ + 2p²q² (3)

Now,

adding 1 and 2, we get

a² + b² =  p⁴ + q⁴ - 2p²q² + 4p²q²

or

a² + b² =  p⁴ + q⁴ + 2p²q²         (4)

equation (3) and (4), we get

a² + b² = c²

⇒ p⁴ + q⁴ + 2p²q² = p⁴ + q⁴ + 2p²q²

x intercept = (-1,0)   (4,0)