X and Y are positive integers which $x^3 + 90\sqrt{xy} + y^3 = 1000$ and $x - \sqrt{xy} + y = 7$; what is the value of $(x^2 + xy + y^2)$?

Question

X and Y are positive integers which $x^3 + 90\sqrt{xy} + y^3 = 1000$ and $x - \sqrt{xy} + y = 7$; what is the value of $(x^2 + xy + y^2)$ !

let's take the second equation :

$x - \sqrt{xy} + y = 7$

squared on both side and get :

$(x - \sqrt{xy} + y)^2 = 7^2$

$x^2 + xy + y^2 - 2x\sqrt{xy} +2xy - 2y\sqrt{xy} = 7$

$x^2 + xy + y^2 - 2\sqrt{xy}(x +\sqrt{xy} - y) = 7$

$x^2 + xy + y^2 - 2\sqrt{xy}(7)= 7$

$x^2 + xy + y^2 = 14\sqrt{xy} + 7$

Up until here, I confused what should I do to solve the problem.

I've also tried to cubed the second equation :

$x - \sqrt{xy} + y = 7$

$(x - \sqrt{xy} + y)^3 = 7^3$

$x^{3}-3x^{2}\sqrt{xy}+6x^{2}y-3y^{2}\sqrt{xy}+6xy^{2}-7xy\sqrt{xy}+y^{3} = 343$

I cannot relate my calculation with the problem. It seems that I made a wrong approach.

Can someone guide me ?

Answer 1

From the second identity you find that $$\sqrt{xy}=x+y-7,\tag{1}$$ which in particular shows that $x+y>7$. Plugging $(1)$ into the first equation yields $$1000=x^3+90\sqrt{xy}+y^3=x^3+90x+90y-630+y^3,$$ from which it follows that $$1630=x^3+90x+90y+y^3=(x+y)(x^2-xy+y^2+90),\tag{2}$$ where $1630=2\times5\times163$. Because $x$ and $y$ are positive we have $$x^2-xy+y^2>x^2-2xy+y^2=(x-y)^2\geq0,$$ and so the second factor in $(2)$ is greater than $90$, so it must be divisible by $163$. It follows that $x+y$ divides $10$. We already saw that $x+y>7$ and so we get $x+y=10$ and hence $x^2-xy+y^2+90=163$. Then $$x^2-xy+y^2=73\qquad\text{ and }\qquad \sqrt{xy}=x+y-7=3,$$ from which it follows that $$x^2+xy+y^2=(x^2-xy+y^2)+2\cdot\sqrt{xy}^2=73+2\cdot3^2=91.$$

Answer 2

$$x+y=\sqrt{xy}+7.(*)$$

$$x^3+y^3=-90\sqrt{xy}+1000.(**)$$

$$90(*)+(**):90x+90y+x^3+y^3=1630. $$

$$\implies (x+y)(x^2-xy+y^2+90)=1630. $$

$9\le x+y\le 18$, $x\ge y$, We derive successively: $$x+y=10,\quad x^2-xy+y^2=73,\quad x=9,\quad y=1,$$ $$x^2+xy+y^2=91.$$

Answer 3

• $x^3 + 90\sqrt{xy} + y^3 = 1000$ implies $x^3 + y^3$ is even as $90\sqrt{xy}$ and $1000$ are both even. This implies $x^3$ and $y^3$ are both odd or both even. In any case, $x+y$ is necessarily even.
• From the second equation, $\sqrt{xy}$ has to be odd, as $x + y$ is even and $7$ is odd. This implies x and y are both odd, and also, $xy$ is a perfect square as the sum of one rational and one irrational term cannot be rational.
• $(x^3 + y^3) < 1000$ implies $(x + y) < 10$, as $(x^3 + y^3)<(x+y)^3$ for all positive integers $x, y$ , which can be seen by the expansion for $(x+y)^3$. Therefore, $x<10$ and $y<10$.

From the above three points, we know that:

1. $x$ and $y$ are less than $10$.
2. $x$ and $y$ are both odd.
3. The product of $x$ and $y$ is a perfect square.

The pairs of integers that satisfy the above conditions are $\{1, 1\}, \{3, 3\},\{5, 5\}, \{7, 7\}, \{9, 9\},$ and $\{1, 9\}$. However, if $x = y$ then the second equation becomes $x - \sqrt{x^2} + x = 7$, which is equivalent to $x =7$. This leaves us with just two pairs of numbers, $\{7, 7\}$ and $\{1, 9\}$. Out of these, only one pair satisfies the first equation, that is $\{1, 9\}$.

Therefore, $x^2 + {xy} + y^2 = $

$1^2 + 1*9 + 9^2 = $ $91$

Answer 4

Another way.-By symmetry we can assume that $x\le y$ (if $(x,y)$ is solution so is $(y,x)$).

If $x=y$ then $x=7$ what is not admissible and $x,y\lt 10$ because $10^3=1000$. On the other hand, $xy$ is a square so we have $$(x,y)=(1,9)\text { or }(1,4)\\$$ with $x=2,3,4,5$ the cases for $xy$ square are with $y=x$. Thus $(x,y)=(1,9)$.