Question

Find two whole numbers, such that the differnece of their squares and the difference of their cubes is a square. What is the answer of the smallest numbers?

Answer 1

We look for whole numbers x,y such that there exists whole numbers z, w such that
\[ x^2-y^2 = z^2\]
and \[x^3-y^3=w^2 \]

Answer 2

i don't understand how i am supposed to even start, other than using that equation.

Answer 3

Note that \[ x^2 = z^2 + y^2 \] is a pythagorean triple. There is a parametric solution to that.

Answer 4

So if u,v are whole numbers then \[ \begin{array}{l} x=u^2+v^2\\y=2uv\\z=u^2-v^2\end{array} \] are such that (x,z,y) is a pythagorean triple. Now we can plug that into the second equation and get \[ (u^2+v^2)^3 - 8u^3v^3 = w^2 \]

Answer 5

The expression on the left can be factored into
\[ \left( {u}^{4}+2\,v{u}^{3}+6\,{u}^{2}{v}^{2}+2\,u{v}^{3}+{v}^{4}
\right) \left( u-v \right) ^{2} \]

Answer 6

We want that to be a square. To make things easier, we set v=1.
As the right factor is already square, we only need that \[ u^4+2u^3+6u^2+2u+1 \] is square.

Answer 7

Now try some small numbers to see if we are lucky. And indeed, u=4 gives 196 which is 14^2.

Answer 8

So we change that back into x and y, then we get x=10 and y=6.
Checking we see that \[ \begin{array}{l}10^2-6^2=8^2\\10^3-6^3=28^2\end{array}\]
Note, that this is just ONE solution. There may be infinitely many! Try finding all of them if you like a challenge.

Answer 9

Thank you so much!

Answer 10

Please hit "Good answer" on one of the answers so that everybody knows, this question is not open anymore.