x and y are two positive integers such that x^2 + y^2 = 1753 and x^3 + y^3 = 52451. Find x + y.

Thanks for helping!

Guest Sep 1, 2020

#2**+2 **

A binomial in the form a3 + b3 can be factored as (a + b)(a2 – ab + b2).

\(x^3+y^3=(x+y)(x^2+y^2-xy)\\ 2(x^3+y^3)=(x+y)(2x^2+2y^2-2xy)\\ \frac{2(x^3+y^3)}{(x+y) }=2x^2+2y^2-2xy\quad(1)\\~\\ (x+y)^2=x^2+y^2+2xy \qquad(2)\\ (1)+(2)\\ \frac{2(x^3+y^3)}{(x+y) }+(x+y)^2=3x^2+3y^2\\ 2(x^3+y^3)+(x+y)^3=3(x+y)(x^2+y^2)\\ Let \;\;A=x+y\\ 2(x^3+y^3)+A^3=3A(x^2+y^2)\\ 2*52451+A^3=3A*1753\\ 104902+A^3=5259A\\ A^3-5259A+104902=0\\\)

Wolfram|Alpha factored this as

\((A-59)(a^2+59A-1778)=0\)

**So A can be 59, this will be the only integer value of A**

so

\(\fbox{x+y=59} \)

LaTex:

x^3+y^3=(x+y)(x^2+y^2-xy)\\

2(x^3+y^3)=(x+y)(2x^2+2y^2-2xy)\\

\frac{2(x^3+y^3)}{(x+y) }=2x^2+2y^2-2xy\quad(1)\\~\\

(x+y)^2=x^2+y^2+2xy \qquad(2)\\

(1)+(2)\\

\frac{2(x^3+y^3)}{(x+y) }+(x+y)^2=3x^2+3y^2\\

2(x^3+y^3)+(x+y)^3=3(x+y)(x^2+y^2)\\

Let \;\;A=x+y\\

2(x^3+y^3)+A^3=3A(x^2+y^2)\\

2*52451+A^3=3A*1753\\

104902+A^3=5259A\\

A^3-5259A+104902=0\\

(A-59)(a^2+59A-1778)=0

Melody Sep 2, 2020