Area of Square possibility

Question

The area of a square is given to be equal to $x^6 \pm x^5 \pm x^4 \pm x^3 \pm x^2 \pm x \pm 1$. Is such an area possible such that it would be equal to a whole number??? If yes, what would be an expression for the area???
I do not have any idea how to approach the problem.
Please Help.
Thankyou.

Answer 1

For each choice of $\pm$, $x^6 \pm x^5 \pm \ldots \pm 1$ is a continuous function and unbounded above. By the Intermediate Value Theorem it takes every real value $\ge 1$, including the squares of all nonzero integers.

It would be a more interesting question if you required $x$ to be an integer, or at least rational. According to Maple, each of the curves $x^6 \pm x^5 \pm \ldots \pm 1 - y^2$ has genus $2$. By Faltings's theorem, they have only finitely many rational solutions.

Answer 2

Here's some I found with a brute force search.

$1+7+7^2-7^3-7^4+7^5+7^6==363^2$
$1+9-9^2+9^3-9^4-9^5+9^6==683^2$
$1-3-3^2-3^3+3^4-3^5+3^6==23^2$
$-1-2+2^2+2^3+2^4+2^5+2^6==11^2$
$-1-2-2^2+2^3-2^4+2^5+2^6==9^2$
$-1-2-2^2+2^3+2^4-2^5+2^6==7^2$
$-1-2+2^2+2^3-2^4-2^5+2^6==5^2$
$-1-2+2^2-2^3-2^4-2^5+2^6==3^2$
$-1-2-2^2-2^3-2^4-2^5+2^6==1^2$

Mathematica code

Select[Flatten[Table[With[{gg = Total[((-1)^# & /@ IntegerDigits[pp, 2, 7])
Table[x^n, {n, 0, 6}]]}, {IntegerQ[Sqrt[gg]], gg, pp, x }], {pp, 0,
 127}, {x, 2, 20}], 1], First[#] == True &]

Fifth powers have two general solutions.

$1 - (n^2 -1) + (n^2 -1)^2 + (n^2 -1)^3 - (n^2 -1)^4 + (n^2 -1)^5 == (n(3 - 3 n^2 + n^4))^2$
$-1 - (n^2 +1) - (n^2 +1)^2 + (n^2 +1)^3 + (n^2 +1)^4 + (n^2 +1)^5 == (n(3 + 3 n^2 + n^4))^2$

Seventh powers have two general solutions.

$ 1 - (n^2 - 1) + (n^2 - 1)^2 - (n^2 - 1)^3 - (n^2 - 1)^4 + (n^2 - 1)^5 - (n^2 - 1)^6 + (n^2 - 1)^7 == (n (-4 + 6 n^2 - 4 n^4 + n^6))^2$

$-1 - (n^2 + 1) - (n^2 + 1)^2 - (n^2 + 1)^3 + (n^2 + 1)^4 + (n^2 + 1)^5 + (n^2 + 1)^6 + (n^2 + 1)^7 == (n (4 + 6 n^2 + 4 n^4 + n^6))^2$

If there is a general solution for power 6, it will likely be findable by a study of the solutions, but so far the solutions seem to be sporadic.

Answer 3

If x can be any real then $x^6 \pm x^5 \pm x^4 \pm x^3 \pm x^2 \pm x \pm 1$ = $\sqrt{x^6 \pm x^5 \pm x^4 \pm x^3 \pm x^2 \pm x \pm 1}$ and the only criterion is the almost trivial $x^6 >= - other terms.

If x must be a positive integer or the length must be rational or some such requirement that $x^6 \pm x^5 \pm x^4 \pm x^3 \pm x^2 \pm x \pm 1$ must be factorable. The $x^6 \pm x^5 \pm x^4 \pm x^3 \pm x^2 \pm x \pm 1$ = $(x^2 + something \pm 1)^2$. Then you have $something^2 + 2*something = \pm x^5 \pm x^4 \pm x^3 \pm x^2 \pm x $. Well this is impossible for variable x (as the highest power would need to be even).

Perhaps the simplest answer is $2^6 - 2^5 - 2^4 - 2^3 -2^2 - 2^1 - 1 = 1$.

====

Ah. I see if we specify that x and the area must both be integers how can we find all answers? Well that's a very interesting question....

Answer 4

Here a family of solutions with the integers $(a,b)$. Take the quadratic number $x=a+b \sqrt m$ where $a,b,m\in\mathbb Z$; $m$ square-free. If you want to restrict $x$ to be real you have to make $m>0$, of course. Hence we have $$ x=a+b \sqrt m$$ $$x^2=A_2+B_2\sqrt m$$ $$x^3=A_3+B_3\sqrt m$$ $$.......$$ $$x^6=A_6+B_6\sqrt m$$ All solution of $\pm b\pm B_2\pm B_3\pm B_4\pm B_5+ B_6=0$ gives a couple $(a,b)$ with $$\pm a\pm A_2\pm A_3\pm A_4\pm A_5+ A_6\pm 1$$ for which one has $$x^6\pm x^5\pm x^4\pm x^3\pm x^2\pm x\pm 1$$ is an integer.You must discard the solutions for which this integer is negative.

Similar way can be tried, (more difficult to manipulated obviously if degree greater than $2$) for $x$ algebraic number.