Squares and Cubes.

Algebra Level 3

There exists two positive whole numbers a and b a \text { and } b such that the difference of their squares is a perfect cube and the difference of their cubes is a perfect square as follows:

{ a 2 − b 2 = c 3 a 3 − b 3 = ( 7 c ) 2 c > 0 \large\begin{cases} a^2-b^2=c^3\ \\ a^3-b^3=(7c)^2\ \\ c>0\end{cases}

What is the minimum of a + b = ? \large a+b=?


The answer is 16.

Hana Wehbi by Hana Wehbi , 51, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Hana Wehbi
Nov 17, 2018

Note that: 1 0 2 − 6 2 = 64 = 4 3 10^2-6^2= 64 = 4^3 and 1 0 3 − 6 3 = 784 = ( 28 ) 2 10^3-6^3= 784 = (28)^2 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

(a^2+ab+b^2)/(a+b)=49/c Let a+b=pc, where p is a positive real number. Then a^2+ab+b^2=49p. This implies a^2+a(pc-a)+(pc-a) ^2=49p Or a^2-apc+a^2-49p=0. Therefore 196p-3p^2.c^2 must be a perfect square. This condition holds for p=4. Then a=6 and b=4 gives a solution. (I tested the case of p=1, which gave no solution. The cases p=2 and p=3 may be tested!)

A Former Brilliant Member - 2 years, 5 months ago

Log in to reply

Makes sense, l would write this as a solution.

Hana Wehbi - 2 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...