Sum of Squares and a Product

Algebra Level 5

Let $N$ be the product of all possible values of $x$ such that $(x,y)$ satisfies the equation $x^2+y^2=28-xy$ , and that both $x$ and $y$ are integers. Find the last three digits of $|N|$ .

The answer is 304.

6 solutions

Edmund Heng
Dec 17, 2013

Rearranging the equation in terms of $y$ , $y^2+(x)y+(x^2-28)=0$ $y=\frac{-x \pm \sqrt {x^2 -4(1)(x^2-28)}}{2(1)}$ $y=\frac{-x \pm \sqrt {112-3x^2}}{2}$ Now we need to fulfill 3 conditions :

To get an integer $y$ , $112-3x^2$ must be a perfect square
For real values of $y$ , $112-3x^2$ must be $\geq 0$ .
The maximum value of $112-3x^2$ is 112 for any value of $x$ , so the perfect square must be $< 112$

$\Rightarrow 0 \leq 112-3x^2 < 112$ where $112-3x^2$ is a perfect square.

Perfect squares within the range are $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$ . Remember that $x$ must be an integer too! Let $k=112-3x^2$ where $k$ is the perfect square listed, $x^2 = \frac {112-k}{3}$ $112-k$ must be divisible by 3, the only $k$ 's that fulfill are $4, 64, 100$ . Solving each of them we get $x = \pm 6, \pm 4, \pm 2$ $\Rightarrow |N| =|6^2*4^2*2^2|=2304$ Last 3 digits is $304$

Wait dude!? I think there is a mistake in this problem. Your explanation is good, but you forget that for every value of $\,x$ you will get 2 distinct values of $\,y$ . For example, if $\,x=-2$ , then $y_1=\frac{-(-2)+\sqrt{112-3(-2)^2}}{2}=6$ and $y_2=\frac{-(-2)-\sqrt{112-3(-2)^2}}{2}=-4.$ Thus, the answer must be $\boxed{416}$ since $|N|=\left|(-2)^2\cdot (2)^2\cdot (-4)^2\cdot (4)^2\cdot (-6)^2\cdot (6)^2\right|=5308416.$ I wanna claim my point back! And also give me extra points for the right answer. $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Tunk-Fey Ariawan - 7 years, 4 months ago

I've just plotted $x^2+y^2+xy=28$ in R and I get a geometric figure ellipse. That strongly indicates that for every value of $\,x$ , you will get 2 distinct values of $\,y$ .

Tunk-Fey Ariawan - 7 years, 4 months ago

Yes u are right, but the question is asking for possible values of $x$ only, which means repeated ones are counted once only. Hope this clears up a lil of ur confusion here.

Edmund Heng - 7 years, 4 months ago

@Edmund Heng – But the equation itself, has 12 distinct combination integers as the solution. So, in my point of view, it has 12 values of $\,x$ .

Tunk-Fey Ariawan - 7 years, 4 months ago

@Tunk-Fey Ariawan – They are asking you the product of the "possible" values of "x", so this solution is absolutely correct.

Shourya Pandey - 7 years, 3 months ago

right...416 should be the answer

Alok Sharma - 7 years, 3 months ago

Agreed...I also did the same thing......

Biswadeep Sen - 7 years, 3 months ago

The solution is very simple, for every x you have two values of y , say {p, q} , such that-
x + p + q = 0 and x takes values {-6, -4, -2, 2, 4, 6} , so you can multiply each x and i think its 2304 not 5308416 .

Abhimanyu Singh - 7 years, 2 months ago

I missed on interpretation of values of x and got 5308416 =2304^2.

My solution:-

$x^2+y^2=28-xy~~\implies~~(x+y)^2=28+xy.~~So~RHS~also~is~a~square.\\ ~~\\ To~ make~RHS~a~perfect~square,~~we~should~have~the~xy~values~given~in~the~column~2~of~following~table. \\ Only~if,~for~a~row~we~ get~same~value~in~column~1~and~4,~x~and~y~are~the~solution. \\ Since~x~and~y~are~interchangeable, ~order~does~not~matter~and~what~is~true~for~y~is~also~true~for~x. \\ Two~factors~of ~xy~~we~select~were~where~x+y~is~smallest.~Only~this~ pair~may~be~in~the~solution~set.~In~other~pairs~LHS~> ~RHS.\\ ~~~\\ ~~~\\ \begin{array}{c|c|c|c||c|||c}We~ Want & & & We~ get & Remarks\\ (x+y)^2 & xy & x~~,~~ y & (x+y)^2 & \\ \hline 1 & -27 & \pm 3 ,~\mp 9 & 36 &\\ 4 & -24 & \pm 4 ,~\mp 6 & 4 & Good...x=\pm~4~~and~~\mp~6\\ 9 & -19 & \pm 1 ,~\mp 19 & 18^2 &\\ 16 & -12 & \pm 2 ,~\mp 6 & 16 &Good...x=\pm~2~~and~~\mp~6\\ 25 & -3 & \pm 1 ,~\mp 3 & 4&\\ 36 & 8 &\pm 2 , ~\pm 4& 36 &Good...x=\pm~2~~and~~\pm~4\\ 49 & 21 & \pm 3 ,~\pm 7 &100 &\\ Bigger~than~49 & & & & LHS~>~RHS\\ \end{array}\\ ~~~\\ ~~~\\ \therefore~product~of~POSSIBLE~x~=~(2*(-2))*(4*(-4))* (-6*6)=~-2304\\ So~~|N|=\Large~~\color{#D61F06}{304}$

Niranjan Khanderia - 3 years, 1 month ago

Jeffrey Robles
Dec 16, 2013

Let $x$ and $y$ be the roots of a quadratic equation.

Also, let $s=x+y \\ p=xy$

If $x$ and $y$ are integers, then $s$ and $p$ are integers too.

Rewriting the given relationship, we have $p=s^2-28$ Then, the quadratic equation whose roots are $x$ and $y$ is $: \\ \qquad \qquad \qquad \qquad k^2-sk+(s^2-28)=0$ .

We solve for the roots of such quadratic equation $\qquad \qquad \quad \qquad \qquad k=\frac{s\pm \sqrt{s^2-4s^2+112}}{2}=\frac{s\pm \sqrt{112-3s^2}}{2}$

W.L.O.G, consider $x>y\\ \quad x=\frac{s + \sqrt{112-3s^2}}{2}; \quad y=\frac{s - \sqrt{112-3s^2}}{2}\\ \quad 112-3s^2\geq0 \\ \quad s \in \{-6,-5,-4,\ldots,4,5,6\}$

Substituting possible values of $s$ in the expression for $x$ and $y$ , we see that $x$ and $y$ are both integers only when $s=\pm2,\pm4,\pm6$ .

When $s=2, x=6, y=-4 \\ s=-2, x=4, y=-6 \\ s=4, x=6, y=-2 \\ s=-4, x=2, y=-6 \\ s=6, x=4, y=2 \\ s=-6, x=-2, y=-4$

Hence there are 6 possible values for $x$ , namely: $\pm2,\pm4,\pm6$ . $|N|=|-2^2 \cdot -4^2 \cdot -6^2|=2304$

you considered $x>y$ but without this you get 6 new results, just by swapping x and y. since x>y hasn't been stated, shouldn't we consider them too?

Pouya Hamadanian - 7 years, 5 months ago

Yes, thanks. I forgot to mention that, but such case should be considered indeed. Without considering that, one would obtain only three possible values for $x$ .

Jeffrey Robles - 7 years, 5 months ago

But, when x=0, y=±√28 Like that, you get real value of y for x=±1,±3,±5 also. Substitute and check. You won't get for integer x>6 or x<-6 because then the quadratic in y so formed won't have real solutions.

Meenakshi Janardanan - 7 years, 5 months ago

Dude!! I think there is a mistake in this problem. See my explanation above! Jeez! I wanna claim my point back! And also give me extra points for the right answer.

Tunk-Fey Ariawan - 7 years, 4 months ago

Adit Mohan
Jan 16, 2014

possible solutions are (-2,6),(6,-2),(-6,2),(2,-6),(2,4),(4,2),(-2,-4),(-4,-2).
different values of x are -2,6,-6,2,4,-4.
product = -2304 last three digits of |-2304|= 304

Israel Smith
Dec 19, 2013

Anyone?

Bharat Karmarkar
Dec 17, 2013

step 1: write table of squares. Select x and y integers. compute the value of x^2 + y^2 + xy step 2: Get the result

Diego E. Nazario Ojeda
Dec 17, 2013

The expression can be written as $x^{2} + xy + y^{2} = 28$ . Note that $-5 \leq x, y \leq 5$ . If you check this cases in terms of $y$ (looking for the possibilities for $x$ ), you will get the following quadratic polinomials in $x$ :

$x^{2} ± 5x - 3$

$x^{2} ± 4x - 12$

$x^{2} ± 3x - 19$

$x^{2} ± 2x - 24$

$x^{2} ± x - 27$

$x^{2}$

The only ones with integer roots are the second and the fourth pairs ( $3$ rd , $4$ th , $7$ th and $8$ th ). The roots are $±6$ , $±4$ and $±2$ , whose product is $-2304$ . Thats why the last three digits of $|N|$ are $\boxed {304}$ .

Mistake: $-6 \leq x, y \leq 6$ and not $-5$ and $5$ . This gives a new pair: $x^{2} \pm 6x + 8$ . Fortunately, this doesn't has an integer root. Sorry! :D

Diego E. Nazario Ojeda - 7 years, 5 months ago

Sum of Squares and a Product

The answer is 304.

6 solutions

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