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

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