Thread: AMATYC question

1. AMATYC question

if a^2-b^2=33 and a^3-b^3=817 have the integer solution with a>b, find the value of a-b

2. Can you solve $\displaystyle (33+b^2)\sqrt{33+b^2}-b^3=817$ ?

3. Hello, rai2003!

I expected a clever solution, but couldn't find it.
So I was reduced to TinkerToy steps.

$\displaystyle \text{If }\;\begin{Bmatrix} a^2-b^2& =& 33 & (1)\\ a^3-b^3 &=& 817 & (2)\end{Bmatrix}\;\text{ has an integer solution}$

$\displaystyle \text{ with }a>b,\:\text{find the value of }a-b.$

In equation [1], the difference of two squares is 33.

There are several solutions: .$\displaystyle (a,b) \;=\;(\pm17,\:\pm16),\;(\pm7,\:\pm4)$

But only one satisfies equation [2]: .$\displaystyle (a,b) \,=\,(17,\:16)$

Therefore: .$\displaystyle a-b\:=\:1$

4. Here's a slightly quicker solution. Since $\displaystyle (a-b)(a+b) = a^2-b^2 = 33$, it follows that $\displaystyle a-b$ is a factor of 33, so it must be $\displaystyle \pm1,\ \pm3,\ \pm11$ or $\displaystyle \pm33$. But $\displaystyle a-b$ is also a factor of $\displaystyle a^3-b^3 = (a-b)(a^2+ab+b^2) = 817$. However, 817 is not divisible by 3 or 11, and it follows that $\displaystyle a-b = \pm1$. You then only have a couple of cases to check in order to conclude that a=17 and b=16.

5. thanks