Find all integer $x$ s.t. there is an integer $r$.
$x^2+(x+1)^2=r^2$
What if $x$ is a square?
All of my attempts lead to dead ends.
Asked
Active
Viewed 1,170 times
1
-
1You you know how to solve Pell's equation? In the general case ($x$ square or note) it reduces to that, but if you don't know how to solve that... – Thomas Andrews Jul 11 '14 at 22:46
-
Please show me how it reduces to a Pell equation. – shooting-squirrel Jul 11 '14 at 22:48
-
@ThomasAndrews, sorry, I was too hasty in posting my answer. – PrimeRibeyeDeal Jul 11 '14 at 22:53
-
3$2x^2+2x+1=r^2$ iff $4x^2+4x+2=2r^2$ iff $(2x+1)^2+1=2r^2$. So we are looking at Pellian $z^2-2r^2=-1$. Now general solution can be written down, or a recurrence. – André Nicolas Jul 11 '14 at 22:53
-
No need to apologize, @PrimeRibeyeDeal. We've all done it. – Thomas Andrews Jul 11 '14 at 22:55
4 Answers
5
If you know how to solve Pell's equation, then you are trying to solve:
$$(2x+1)^2-2r^2 = -1$$ which is a common variant of Pell's equation.
Thomas Andrews
- 177,126
-
My answer obviously doesn't answer the question about when $x$ is a square, which is harder... – Thomas Andrews Jul 11 '14 at 22:58
-
Could you recommend me a good refference to solving Pell equations? – shooting-squirrel Jul 11 '14 at 23:12
-
Not other than saying, "google Pell's equation." In this case, it's just: $2x+1+r\sqrt 2 = (1+\sqrt 2)^{2k+1}$. – Thomas Andrews Jul 11 '14 at 23:22
1
About the case where $x$ is square:
Following the standard solution to Pell's equation, the solution to this problem is given by the following recurrences: $$y_n = 2x_n + 1$$ $$y_0 = 1, y_1=7, y_{n+2}=6y_{n+1}-y_n$$ $$r_0 = 1,r_1=5, r_{n+2}=6r_{n+1}-r_n$$
So the problem is equivalent to finding out when $y_n-1$ is equal to twice a square. $n=0$ is probably the only solution. I suspect that this is a moderately hard but solvable problem—in principle it should be approximately as hard as finding all square Fibonacci numbers, which is very tricky to do with elementary methods.
Andrew Dudzik
- 30,074
-
For those interested, there is a closed form: $$ y_n=\frac{1+\sqrt2}2(3+2\sqrt2)^n+\frac{1-\sqrt2}2(3-2\sqrt2)^n $$ – mathreader Nov 17 '17 at 03:04
0
Here's a series of reductions you may find helpful:
Consider the more general problem of finding triplets of integers $(a,b,c)$ such that
$$a^2 + b^2 = c^2$$
It's know that if $a = m^2-n^2$and $b=2mn$ that in general
$$(m^2-n^2)^2 + (2mn)^2 = m^4+ 2m^2n^2 + n^4= (m^2+n^2)^2$$
Thus for every pair of integers $m,n$ we can generate a triple $a,b,c$ that satisfies this equation.
Now we are concerned with solutions such that $2mn = m^2-n^2 + 1$
Sidharth Ghoshal
- 16,771
0
It is clear that there is a relationship with solutions of Pell's equation, but it must still be written more clearly and in a more general way.
Solutions of the equation: $$X^2+(X\pm{a})^2=Y^2$$
Defined solutions of Pell's equation: $$p^2-2s^2=\pm{a}$$
And the solutions are of the form: $$X=2s(s+p)$$ $$Y=2s(s+p)+p^2$$
These numbers can be different characters. I almost forgot.
If we know what the solution of Pell's equation $$p^2-2s^2=a$$ $$s_{2}=2p+3s$$ $$p_{2}=3p+4s$$
individ
- 4,301