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Today, I came across an equation in practice mock-test of my coaching institute, aiming for engineering entrance examination (The course for the test wasn't topic-specific, it was a test of complete high school mathematics). It was having four variables and only one equation. While analyzing my test paper, this is the only problem I (and my friends too) couldn't figure out even after giving this problem several hours. So I came here for some help.

Question : Solve for $a,b,c,d \in \Bbb R$, given that $$a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$$

Since only one equation is given, there must be involvement of making of perfect squares, such that they all add up to $0$. Thus, resulting in few more equations. But how to?

I tried a lot of things, such as making $(a-b)^2 $ by adding the missing terms and subtracting again, but got no success.

Thanks!

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    $\begingroup$ Is it really "$... -d + \frac{2}{5}$" instead of "$...-da +\frac{2}{5}$"? $\endgroup$ – achille hui Nov 9 '17 at 14:23
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    $\begingroup$ @achillehui Yes. The equation is exactly what it's written. $\endgroup$ – Jaideep Khare Nov 9 '17 at 14:28
  • $\begingroup$ The equation is equivalent to $(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + 2ad - 2d + \frac{4}{5} = 0$, which at least put some restrictions on $a$ and $d$ (e.g., if $d$ is positive then $a$ is less than 1). $\endgroup$ – Connor Harris Nov 9 '17 at 14:38
  • $\begingroup$ $(a,b,c,d) = (1/5,2/5,3/5,4/5)$ $\endgroup$ – achille hui Nov 9 '17 at 14:45
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Your idea of turning the right side into a sum of perfect squares is a good one. Observing that $$ (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=2(a^2+b^2+c^2+d^2-ab-bc-cd-da) $$ we multiply $$ a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac{2}{5}=0 $$ by $2$ and rewrite the result as $$ (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2+2ad-2d+\frac{4}{5}=0. $$ We define $x=a-b$, $y=b-c$, and $z=c-d$, which gives $a=x+y+z+d$. Substituting gives $$ x^2+y^2+z^2+(x+y+z)^2+2d(x+y+z+d-1)+\frac{4}{5}=0, $$ which can be rewritten as $$ x^2+y^2+z^2+(x+y+z+d)^2+d^2-2d+\frac{4}{5}=0 $$ or $$ x^2+y^2+z^2+(x+y+z+d)^2+(d-1)^2=\frac{1}{5}. $$ So a sum of five perfect squares equals $\frac{1}{5}$. We might hope for a solution in which each of the perfect squares is $\left(\pm\frac{1}{5}\right)^2$, and indeed we find that $x=y=z=-\frac{1}{5}$, $d=\frac{4}{5}$ provides such a solution.

Now we ask whether the solution can be perturbed. Letting $x=-\frac{1}{5}+e$, $y=-\frac{1}{5}+f$, $z=-\frac{1}{5}+g$, $d=\frac{4}{5}+h$ and substituting, we get $$ (-1/5+e)^2+(-1/5+f)^2+(-1/5+g)^2+(1/5+e+f+g+h)^2+(-1/5+h)^2=\frac{1}{5}, $$ which simplifies to $$ e^2+f^2+g^2+h^2+(e+f+g+h)^2=0. $$ This forces $e=f=g=h=0$, and therefore the solution is unique.

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  • $\begingroup$ Amazing! This is the one solution that is practically possible to think during examination. Thanks for this!😊 $\endgroup$ – Jaideep Khare Apr 7 '18 at 13:30
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Let $F(a,b,c,d) = a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac25$.

With help of a CAS, one can verify

$$\begin{align} F\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right) &= p^2 - pq + q^2 - qr + r^2 -rs + s^2\\ &= \frac12\left(p^2 + (p-q)^2 + (q-r)^2 + (r-s)^2 + s^2\right)\end{align}$$

If one set $(a,b,c,d)$ to $\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right)$, one find $$\begin{align} F(a,b,c,d) = 0 &\iff p = p-q = q-r = r-s = s = 0\\ &\iff p = q = r = s = 0 \end{align} $$ This implies the equation at hand has a unique solution: $$(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$$

Update

About the question how I come up with this. I first write $F(a,b,c,d)$ as

$$\begin{align} F(a,b,c,d) &= a^2 + b^2 + c^2 + d^2 - ab - bc - cd - da + d(a-1) + \frac25\\ &= \frac12((a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2) + d(a-1) + \frac25 \end{align}\tag{*1} $$ To simplify the term $d(a-1)$, I introduce $\lambda, \mu$ such that

$$\begin{cases} d &= \frac12 + \lambda + \mu\\ a &= \frac12 + \lambda - \mu \end{cases} \quad\implies\quad d(a-1) = \lambda^2 - \left(\frac12+\mu\right)^2 $$ Now $d-a = 2\mu$ and $(a-b)^2 + (b-c)^2 + (c-d)^2 \ge 3\left(\frac{d-a}{3}\right)^2 = \frac43 \mu^2$.

If one substitute this back into $(*1)$, one find

$$F(a,b,c,d) \ge \frac83\mu^2 + \lambda^2 - (\frac12 + \mu)^2 + \frac25 = \lambda^2 + \frac53\left(\mu - \frac{3}{10}\right)^2$$

In order for $F(a,b,c,d) = 0$, we need

$$\lambda = 0,\quad\mu = \frac{3}{10} \quad\text{ and }\quad(a-b)^2 + (b-c)^2 + (c-d)^2 = \frac13(d-a)^2$$ The last condition forces $d-c = c-b = b-a = \frac13(d-a)$ and leads to the solution $(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$. This is a little bit sloppy to describe, so I look at expansion of $F(a,b,c,d)$ near the solution and obtain a simpler description of $F$ in terms of $p,q,r,s$.

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  • $\begingroup$ Amazing! (+1)BTW, How did you come up with this solution, I mean how did you think about doing so? $\endgroup$ – Jaideep Khare Nov 9 '17 at 15:17
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    $\begingroup$ @JaideepKhare see update in answer. $\endgroup$ – achille hui Nov 9 '17 at 15:42
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Multiply by $2$ and rearrange to \begin{align*}(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + 2ad - 2d + \frac{4}{5} = 0. \tag{$\star$}\end{align*} For fixed $a$ and $d$, the minimum value of $(a-b)^2 + (b-c)^2 + (c-d)^2$ is $\frac{(d - a)^2}{3}$, with equality so the LHS of $(\star)$ is at least $$\frac{4}{3} (d-a)^2 + 2ad - 2d + \frac{4}{5} = \frac{4}{3} \left( a - \frac{d}{4} \right)^2 + \frac{5}{4} \left( d - \frac{4}{5}\right)^2 \tag{$\dagger$}.$$ But $(\dagger)$ is clearly non-negative, and it is zero if and only if $d = 4/5$ and $a = d/4 = 1/5$, but the LHS of $(\star)$ must be zero. From this, $b = 2/5$ and $c = 3/5$ follow, and there can be no other solution.

Lemma. For fixed $x_0$ and $x_n$, the sum $\sum_{i=1}^n (x_i - x_{i-1})^2$ is minimized when the $x_i$ form an arithmetic progression $x_i = \frac{n-i}{n} x_0 + \frac{i}{n} x_n$.

Proof. For $n = 2$, $(x_0 - x_1)^2 + (x_1 - x_2)^2$ can be rearranged as $$ \left( x_1 - \frac{x_0 + x_2}{2} \right)^2 +x_0^2 + x_2^2 - \frac{(x_0 + x_2)^2}{4}. $$ For $n > 2$, if some $x_k$ is not the midpoint of $x_{k-1}$ and $x_{k+1}$, then $(x_k - x_{k-1})^2 + (x_{k+1} - x_k)^2$ can be reduced by moving $x_k$ to the midpoint, leaving the other terms of $\sum_{i=1}^n (x_i - x_{i-1})^2$ alone. So if a minimum exists, it must have evenly spaced $x_i$. And proving that a minimum exists is simple: the possible values of $x_1, \ldots, x_{n-1}$ that can minimize $f(x_1, \ldots, x_{n-1}) = \sum_{i=1}^n (x_i - x_{i-1})^2$ can be bounded in some closed interval $[-R, R]$, and the image of a connected compact set $[-R, R]^n$ under a continuous function $f$ must be compact and connected (that is, a closed bounded interval).

This lemma can be interpreted physically as stating that the potential energy of a chain of $n$ identical springs with unstretched length zero, with the endpoints of the whole chain anchored, is minimized (and thus the forces at each spring endpoint are in equilibrium) when each spring is stretched equally. Here, $x_0$ and $x_n$ are the fixed endpoints, and $x_{i-1}$ and $x_i$ are the endpoints of the $i$th spring.

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It's $$a^2-ab+\frac{b^2}{4}+\frac{3}{4}b^2-bc+\frac{1}{3}c^2+\frac{2}{3}c^2-cd+\frac{3}{8}d^2+\frac{5}{8}d^2-d+\frac{2}{5}=0$$ or $$\left(a-\frac{b}{2}\right)^2+\frac{3}{4}\left(b-\frac{2c}{3}\right)^2+\frac{2}{3}\left(c-\frac{3d}{4}\right)^2+\frac{5}{8}d^2-d+\frac{2}{5}=0$$ or $$\left(a-\frac{b}{2}\right)^2+\frac{3}{4}\left(b-\frac{2c}{3}\right)^2+\frac{2}{3}\left(c-\frac{3d}{4}\right)^2+\frac{5}{8}\left(d-\frac{4}{5}\right)^2=0,$$ which gives the answer.

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  • $\begingroup$ Can you please explain what made you bring the equation in the perfect square form $\endgroup$ – user471651 Nov 10 '17 at 9:13
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    $\begingroup$ @user471651 I added something. See now. $\endgroup$ – Michael Rozenberg Nov 10 '17 at 10:25
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The expression is a quadratic form. Experimenting a bit you can put $v^{T}=(a, b, c, d, 1)$ and let A be the 5 by 5 matrix with 1, 1, 1, 1, $\frac{2}{5}$ down the lead diagonal, with -1, -1, -1, -1 above that and 0 everywhere else. Then the expression is $v^{T}Av = 0$. The expression is also $v^{T}A^{T}v = 0$ and better still it is $v^{T}(A+A^{T})v = 0$ where $A+A^{T}$ is a symmetric matrix. In this form $(A+A^{T})v = 0$ has the unique solution v already given. It would be nice if it were obvious that this is the unique solution of $v^{T}(A+A^{T})v = 0$ but I can't see it.

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I think that the following is most natural way to solve this equation. $$a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$$ Since it is quadratic on $a$ we have nonegative discriminant:

$$ b^2- 4(b^2+c^2+d^2-bc-cd-d+\frac 25) \geq 0$$ so $$4c^2-4c(b+d)+4d^2+3b^2-4d +\frac85 \leq 0$$

so $$4c^2-4c(b+d)+(b+d)^2-(b+d)^2+4d^2+3b^2-4d +\frac85 \leq 0$$ so $$(2c-b-d)^2+3d^2+2b^2-2bd-4d+\frac85 \leq 0$$ so
$$3d^2+2b^2-2bd-4d+\frac85 \leq 0\;\;\;/\cdot 2$$ so $$4b^2-4bd+d^2+5d^2-8d+\frac{16}5 \leq 0 \;\;\;/\cdot 5$$ so $$5(2b-d)^2+25d^2-40d+16 \leq 0 $$ so $$5(2b-d)^2+(5d-4)^2 \leq 0 $$ which means that $d=4/5$ and $b=2/5$ and $c=(b+d)/2=...$ and $a=...$

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