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

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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!

Sep 1, 2020

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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)\\

(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

Sep 2, 2020