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


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.

  • $\begingroup$ 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

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.


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