We have received two good solutions to this problem.

Biren Patel (aged 14) from The Heathland School approached it like this:

The question asks you to prove that a+b=c.

You already know that the two lines are parallel.This tells us that they have the same gradient.

To work out the gradients we should try to imagine we draw a right angled triangle under both lines.

We then divide the change in the y-axis by the change in the x-axis.

The change in the x-axis of the line AB is equal to b-a (small letters refer to the co-ordinates). The change in the y-axis is equal to

${\displaystyle b^2-a^2}$

(these letters also refer to the co-ordinates).

Thus the gradient of line AB is equal to:

${\displaystyle \frac{b^2-a^2}{b-a}=\frac{(b-a)(b+a)}{b-a}=b+a}$

The change in the y-axis of the line OC is $\hspace{1em}{c}^2.$ The change in the x-axis is equal to c. Thus the gradient of the line OC is

${\displaystyle \frac{c^2}{c}=c}$

We know that both lines, AB and OC are parallel, and so they must have the same gradient.

So b+a = c

Andrei Lazanu (aged 12) from School No. 205 in Bucharest (Romania), approached it in a slightly different way:

Let y = mx + n be the equation of the line going through the points A(a, a ²) and B(b, b ²).

At A,

${\displaystyle a^2=\mathrm{ma}+n}$

At B,

${\displaystyle b^2=\mathrm{mb}+n}$

From this:

${\displaystyle a^2-b^2=\mathrm{ma}-\mathrm{mb}=m(a-b)}$

Dividing by (a-b):

${\displaystyle m=\frac{a^2-b^2}{a-b}=\frac{(a+b)(a-b)}{a+b}=a-b}$

The line OC passes through the origin and the point C(c, c ²).

Because it passes through the origin, its equation is of the form:

${\displaystyle y=\mathrm{ux}}$

Because it is parallel to the line passing through A and B, u = m, so its equation is:

${\displaystyle y=(a+b)x}$

Since it passes through C:

${\displaystyle c^2=(a+b)c}$

or:

${\displaystyle c=a+b}$

Well done to you both.