$$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$

I tried many column operations, mainly subtractions without any success.

  • 7
    $\begingroup$ Who comes up with a problem like this?? $\endgroup$
    – imranfat
    Commented Oct 4, 2016 at 18:34
  • $\begingroup$ If I get a constant in a column can it help somehow? $\endgroup$
    – Zauberkerl
    Commented Oct 4, 2016 at 18:38
  • 17
    $\begingroup$ This exercise can be found in Golan's book It is Exercise 641 in the 2012 edition and Exercise 584 in the 2007 edition. (BTW it is good to add also source of the problem when posting the question.) $\endgroup$ Commented Oct 5, 2016 at 4:22
  • 1
    $\begingroup$ This is also a particular case of Exercise 36 (a) in my Notes on the combinatorial fundamentals of algebra ( web.mit.edu/~darij/www/primes2015/sols.pdf ) in the version of 6 October 2016. (If the numbering changes, see the frozen version of 4 September 2016, at github.com/darijgr/detnotes/releases/tag/2016-09-04 .) Exercise 36 (a) is, in turn, a particular case of Exercise 36 (c). $\endgroup$ Commented Oct 7, 2016 at 22:39

7 Answers 7


If you expand the binomials and you subtract the $1^{\rm st}$ column from the 2nd, 3rd, and 4th you get $$ \begin{vmatrix}a^2 & 2a+1 & 4a+4 & 6a+9 \\ b^2 & 2b+1 & 4b+4 & 6b+9 \\ c^2 & 2c+1 & 4c+4 & 6c+9 \\ d^2 & 2d+1 & 4d+4 & 6d+9\end{vmatrix}. $$ Next subtract 2 times the 2$^{\rm nd}$ column from the 3$^{\rm rd}$ one, and 3 times the 2$^{\rm nd}$ column from the 4$^{\rm th}$ one: $$ \begin{vmatrix}a^2 & 2a+1 & 2 & 6 \\ b^2 & 2b+1 & 2 & 6 \\ c^2 & 2c+1 & 2 & 6 \\ d^2 & 2d+1 & 2 & 6\end{vmatrix}=0. $$ Now it is clear that the determinant is zero, as the fourth column is a multiple of the third one (or, factor 3 from the fourth column to get two equal columns).


Hint. Is it possible to find $A,B,C$ such that for all $x\in\mathbb{R}$, $$(x+3)^2=A(x+2)^2+B(x+1)^2+Cx^2?$$

P.S. The answer is yes: $A = 3$, $B = -3$, $C = 1$.

  • 10
    $\begingroup$ And indeed four polynomials of degree two have to be linearly dependent. $\endgroup$
    – Carsten S
    Commented Oct 5, 2016 at 15:28

Switch the sign of columns 2. and 4., then multiply column 2. and 3. by 3 and finally add them up to get zero in every row:

Col.1 - 3 Col.2 + 3 Col.3 - Col.4 = 0

So columns are linearly dependent, hence the determinant is zero for any $a\ldots d$.

  1. Take any sequence of consecutive square numbers. What do you know (or, if you don't know, try it: what do you notice) about the second differences?

  2. Each row of the matrix has four consecutive square numbers.* Note that column operations preserve the determinant in the same way that row operations do (this is obvious if we recall that the determinant of a matrix equals the determinant of its transpose, and column operations on your matrix are row operations on its transpose). One can fill the final three columns of your matrix with the first differences of your four consecutive square numbers, by applying the following column operations: subtract column 3 from column 4; subtract column 2 from column 3; subtract column 1 from column 2. (Why did we have to do it in this order?)

  3. Can you perform another set of column operations to difference these three first differences? The final two columns of your matrix will then be filled with the second differences of your original sequences of four consecutive square numbers. Recall the result from (1). What does this tell you about these two columns and hence the determinant?

$(*)$ Perhaps $a^2, \dots, (a+3)^2$ may not be "square numbers" in the sense that $a$ may not be a natural number. But this doesn't matter very much; the reason the second differences of the sequence $n^2, \, n\in\mathbb{N}$ are so nice is because of algebra that works just as well on $x^2, \, x\in \mathbb{R}$. In particular, what is $\left((x+2)^2-(x+1)^2\right) - \left((x+1)^2-x^2\right)$?

If you brush up a little on finite differences of higher polynomials you might want to have a think about how you could determine the following determinant, where each row has six consecutive fourth powers:

$$\begin{vmatrix}a^4 & (a+1)^4 & (a+2)^4 & (a+3)^4 & (a+4)^4 & (a+5)^4 \\ (b+6)^4 & (b+7)^4 & (b+8)^4 & (b+9)^4 & (b+10)^4 & (b+11)^4 \\ (c-3)^4 & (c-2)^4 & (c-1)^4 & c^4 & (c+1)^4 & (c+2)^4 \\ (d+20)^4 & (d+21)^4 & (d+22)^4 & (d+23)^4 & (d+24)^4 & (d+25)^4 \\ (e-8)^4 & (e-7)^4 & (e-6)^4 & (e-5)^4 & (e-4)^4 & (e-3)^4 \\ (f + 2016)^4 & (f+2017)^4 & (f+2018)^4 & (f+2019)^4 & (f+2020)^4 & (f+2021)^4 \end{vmatrix} $$

Expanding this out in full may not necessarily develop the situation to your advantage. But this time, with a fourth power polynomial, it isn't the second differences that come out the same...

  • $\begingroup$ Don't you want to get zero with a 7x7 matrix in your think? $\endgroup$
    – bobuhito
    Commented Oct 5, 2016 at 0:01
  • $\begingroup$ @bobuhito Thanks. When I wrote it, I realised 7 x 7 wouldn't fit on the screen but I forgot to change the power down from 5 to 4! $\endgroup$
    – Silverfish
    Commented Oct 5, 2016 at 0:23

The determinant is a polynom of the second degree with respect to $a$, on the other hand, it has three different (in general) roots $b$, $c$, and $d$ (as, e.g., if $a=b$, the determinant has two identical rows), therefore, the determinant vanishes.


I wanted to present a different view of the problem that takes advantage of the sameness of the rows in a less computational way.

We can view the entries of each row as the evaluations of 4 polynomials of degree 2: $x^2$, $(x+1)^2$, $(x+2)^2$ and $(x+3)^2$ at a point.

Since the set of polynomials of degree at most 2, $\mathbb{P}_2$, forms a 3 dimensional vector space over $\mathbb{R}$, any collection of 4 polynomials will be linearly dependent. Hence there is some dependence relation among the columns. This dependence relation, $\alpha_0x^2+\alpha_1(x+1)^2+\alpha_2(x+2)^2+\alpha_3(x+3)^2=0$ means some combination of elementary column operations will create an all 0's column. Hence the matrix has linearly dependent columns.


If you expand the squares, you'll see that every column is a linear combination of $$ \begin{bmatrix}a^2 \\ b^2 \\ c^2 \\ d^2\end{bmatrix}, \begin{bmatrix}a \\ b \\ c \\ d\end{bmatrix}, \text{ and } \begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}. $$

There are four columns that are linear combinations of three vectors. So they can't be linearly independent, thus the determinant is $0$.

(Or: The column space of the matrix is spanned by $3$ vectors. So the rank of the matrix is at most $3$, which means the matrix is singular. So the determinant is $0$.)


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