"Consider an 4 term arithmetic sequence. The difference is 4, and the product of all four terms is 585. Write the progression".

My way of finding the progression seems like it will take too long, but here it is, anyway:

$$a_1\cdot a_2\cdot a_3\cdot a_4=585$$

$$a_1\cdot (a_1+4)\cdot (a_1+8)\cdot (a_1+12)=585$$

and after some operations $$a^4 +4a^3+196a^2+384a-585=0 $$

Is there a faster, less frustrating way of solving this? Thanks in advance/

  • 2
    $\begingroup$ Factorise $585$. $\endgroup$ – TonyK Nov 23 '14 at 21:17
  • $\begingroup$ The difference is integer and the product of $a_1,\dots,a_4$ is also integer, hence $a_1,\dots,a_4$ are all integers. $\endgroup$ – barak manos Nov 23 '14 at 21:31
  • $\begingroup$ I don't know how you got that polynomial, but it doesn't look right to me. $\endgroup$ – TonyK Nov 23 '14 at 22:06

The faster way of doing this would be to let your product be $(a - 6)(a-2)(a+2)(a+6) = 585$. This then expands to $(a^2 - 4)(a^2 - 36) = 585$, which substituting $b = a^2$ yields a quadratic which can be more easily factored.

You end up getting $a^2 = 49$ as the only positive root, so you have $a = 7$ and your sequence is: $1, 5, 9, 13$.

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  • $\begingroup$ Amazing, really. But how do I go on about letting my product be that? $\endgroup$ – SharkFin Nov 23 '14 at 21:33
  • $\begingroup$ @SharkFin What do you mean? $\endgroup$ – Tavian Barnes Nov 24 '14 at 0:09
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    $\begingroup$ @SharkFin, you already know that the progression goes from $a_1$ to $a_1 + 12$. All Too_Pro is doing is re-centering at $a = a_1 + 6$, the ``center'' of the progression (even though no element of the progression is there). This allows the constants to be smaller, and we know going left and right the "same amount" must yield numbers in the progression at the same time, by symmetry. $\endgroup$ – Charles Baker Nov 24 '14 at 0:22
  • $\begingroup$ I completely agree with @user52733. Another perk of doing this "shifting" to the center of the progression, in addition to having smaller constants, is that you get the products to become the products of two differences of squares. That's why the substitution works, whereas the OP's attempt yielded a more unwieldy quartic. $\endgroup$ – djsavvy Nov 25 '14 at 0:03

Another way to do it is to factor $585$, which turns out to be $$3^2\cdot 5\cdot 13$$

And using the factorization, find four factors which each differ by $4$.

$$1\times 5\times 9\times 13$$

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  • $\begingroup$ My only qualm with this solution is that it is not very easily generalizable and doesn't actually provide much insight into anything beyond the number 585 or a really bashy way of doing it. However, on a timed competition, this might be the fastest, if not most easily checkable way. $\endgroup$ – djsavvy Nov 25 '14 at 0:01

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