1. ## Cubic Equation

Find all positive integer roots of the equation $1 + n + n^2 + n^3 = A^2$

My solution attempt:
Following Wolframalpha factoring, it can be rewritten as $(n+\frac{1}{3})^3+\frac{2}{3}(n+\frac{1}{3})+\frac {20}{27}=A^2$, then

$\displaystyle{(3n+1)^3+6(3n+1)+20=27A^2$ and putting $3n+1=k \Rightarrow n=\frac{k-1}{3}}$ and for the

first positive integer root, $k_1=4 \Rightarrow n_1=1 \Rightarrow A_1=2$
second positive integer root, $k_2=22 \Rightarrow n_2=7 \Rightarrow A_2=20$

How can I prove that these $[(n, A)=(1, 2),(7, 20)]$ are the only positive integer solutions? Thanks for any suggestions. Regards.

2. ## Re: Cubic Equation

Is (0,1) not a valid solution?

3. ## Re: Cubic Equation

Thanks for your interest. I guess n=0 is unacceptable because it is not considered positive.

4. ## Re: Cubic Equation

Ok.
Solution is also:
n = (u + v)^(1/3) - 2/[9(u + v)^(1/3)] - 1/3
where:
u = SQRT(27A^4 - 40A^2 + 16) / [2(3^(3/2)]
v = (27A^2 - 20) / 54

(0,1) , (1,2) and (7,20) seem to be the only integer solutions with n=>0.
I ran a looper with A from 1 to 999999: only above resulted.
So perhaps above equation can be used to devise the proof...

5. ## Re: Cubic Equation

Thank you very much. I will work on your method. Regards.

6. ## Re: Cubic Equation

Originally Posted by Honore
Find all positive integer roots of the equation $1 + n + n^2 + n^3 = A^2$
You can factorise it as $(1+n)(1+n^2) = A^2.$ The factor $1+n^2$ clearly cannot be a perfect square. So at least one of the squared prime factors in $A^2$ must be shared between the two factors on the left side. But the only possible common factor of $1+n$ and $1+n^2 = (1+n)^2-2n$ is 2. So we can write $n+1 = 2s^2$ and $n^2+1 = 2t^2$ for some natural numbers $s,t.$

Therefore $(2s^2-1)^2 + 1 = 2t^2$ which, with a bit of reorganisation and after cancelling the factor 2, can be written as $(s^2-1)^2 + s^4 = t^2.$ This shows that the set $(s^2-1, s^2, t)$ is a Pythagorean twin triple. There are infinitely many Pythagorean twin triples, but we want one in which the middle element is a perfect square. The triples (0,1,1) and (3,4,5) satisfy that condition, leading to the values n=1 and n=7. The next triples are (20,21,29), (119,120,169) and (696,697,985), but none of those has a perfect square as the middle element. After that, the numbers increase very rapidly, as you can see from the listing at A046090 - OEIS. My guess is that there are no more perfect squares in that sequence, but I have to admit that I don't begin to see how to prove that.

7. ## Re: Cubic Equation

Thanks a lot. That is much better and efficient than what I have been able to do so far. Regards.