3
$\begingroup$

The following problem was posted to usenet forum de.rec.denksport two weeks ago and no progress was made.

Find all positive integers $x$,$y$ satisfying the equation

$$2x^2 - y^{14} = 1$$

$(1,1)$ is a solution and i suspect it is the only one.

I tried some things, e.g. I started from

$$2x^2=1+y^{14}$$

and using the identities

$$1+y^{14}=(y^2+1)(y^{12}-y^{10}+y^8-y^6+y^4-y^2+1)$$ and $$y^{12}-y^{10}+y^8-y^6+y^4-y^2+1=(y^2+1)(y^{10}-2y^8+3y^6-4y^4+5y^2-6)+7$$

it was possible to split the equations in two equations

$$2u^2=y^2+1$$ $$v^2=y^{12}-y^{10}+y^8-y^6+y^4-y^2+1$$ $$\gcd(y^2+1,y^{12}-y^{10}+y^8-y^6+y^4-y^2+1)=1$$

Using this result I could check that there is no solution for small integers ($y$ less than $10^{2000}$) by finding the set of solutions of the first of these equations using http://www.alpertron.com.ar/QUAD.HTM and checking if $y^{12}-y^{10}+y^8-y^6+y^4-y^2+1$ is a square. But I had no clue how to solve the problem.

$\endgroup$
  • $\begingroup$ the second polynomial of degree 12, the value of it is odd for both y=0 and y=1. I remember reading somewhere such polynomials cannot have integral roots. Hence (1,1) is the only solution $\endgroup$ – Bhargav Feb 4 '12 at 17:08
7
$\begingroup$

The equation $v^2=y^{12}-y^{10}+y^8-y^6+y^4-y^2+1$ has only trivial solutions $y=0,1$. Define $f(x)=1+x+x^2+x^3+x^4+x^5+x^6$ so that the right hand side is $f(-y^2)$. For $x>5$ we have $$(16x^3+8x^2+6x+5)^2<256 f(x)< (16x^3+8x^2+6x+6)^2,$$ while for $x<-4$ we get $$(16x^3+8x^2+6x+5)^2<256 f(x)< (16x^3+8x^2+6x+4)^2.$$ Checking the values $-4\leq x\leq 5$ by hand, we see that $f(x)$ is only a square when $x=0,-1$.

Ribenboim's book on Catalan's conjecture has a detailed analysis of the Diophantine equation $v^2=1+x+x^2+\cdots +x^{n-1}$. The only non-trivial solutions are $n=5, x=3$ and $n=4, x=7$.

See also this MSE problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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