Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{{(a + 1)}^2}}&{{{(a + 2)}^2}}&{{{(a + 3)}^2}}\\{{b^2}}&{{{(b + 1)}^2}}&{{{(b + 2)}^2}}&{{{(b + 3)}^2}}\\{{c^2}}&{{{(c + 1)}^2}}&{{{(c + 2)}^2}}&{{{(c + 3)}^2}}\\{{d^2}}&{{{(d + 1)}^2}}&{{{(d + 2)}^2}}&{{{(d + 3)}^2}}\end{array}} \right| $$

It's very stupid desicion for me to expand it by row or column.

Any suggestions?

share|cite|improve this question
1  
One observation : $$a+(a+3)=a+1+(a+2)$$ So, write $$a=A-2D,a+1=A-D,a+2=A+D,a+3=A+2D$$ Also, think of $$(a+r)^2-(b+r)^2$$ has a factor $a-b$ – lab bhattacharjee Jan 15 at 8:30
up vote 6 down vote accepted

HINT

If you subtract the first column from every other column you get $$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{{(2a + 1)}}}&{{{(4a + 4)}}}&{{{(6a+ 9)}}}\\{{b^2}}&{{{(2b + 1)}}}&{{{(4b + 4)}}}&{{{(6b + 9)}}}\\{{c^2}}&{{{(2c + 1)}}}&{{{(4c + 4)}}}&{{{(6c + 9)}}}\\{{d^2}}&{{{(2d + 1)}}}&{{{(4d + 4)}}}&{{{(6d + 9)}}}\end{array}} \right| $$ next subtract 2 times the second column from the third column and 3 times the second column from the fourth column. Can you take it from here?

share|cite|improve this answer

Hint: Let $P(x)=x^2$. The polynomials $P(x+k)$, $k=0,1,2,3$ lies in the vector space of the polynomials of degree $\leq 2$, that is of dimension $3$. Hence there are dependant, and there exists $u,v,w\in \mathbb{R}$ such that $P(x+3)=uP(x)+vP(x+1)+wP(x+2)$ for all $x$.

share|cite|improve this answer
1  
this is a wonderful solution! – Ant Jan 15 at 10:24

By the fourth column (resp. the third column, the second column) subtract the third column (resp. the second column, the first column), you get $\left|\begin{array}{cccc} a^2 & 2a+1 & 2a+3 & 2a+5 \\ b^2 & 2b+1 & 2b+3 & 2b+5 \\ c^2 & 2c+1 & 2c+3 & 2c+5 \\ d^2 & 2d+1 & 2d+3 & 2d+5 \\ \end{array} \right|=0$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.