Let $n-m \ge 2$ be two fixed natural numbers. Are there any nontrivial rational solutions of the equation $$x^{n-m}=(1+t^m)/(1+t^n)$$ for $x$ and $t$? As particular cases the rational solutions of the equations $x^2=(1+t^2)/(1+t^4)$ and $x^2=(1+t^3)/(1+t^5)$ will be interesting.
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$\begingroup$ I think your first example (2,4) asks for coprime integers r and s with $r^4 + s^4$ being twice a square. With the exception of r=s=1 (and maybe one example of Ljunggren), I believe there are no examples. It was shown by Fermat that the sum of two biquadrates is never a square. I think you can find literature supporting my belief. Gerhard "Have Faith In The Impossibilities" Paseman, 2018.01.31 $\endgroup$– Gerhard PasemanCommented Jan 31, 2018 at 21:02
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$\begingroup$ thanks for your comments.how do you convert my problem to Fermat problem? $\endgroup$– mehdi baghalaghdamCommented Jan 31, 2018 at 21:10
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$\begingroup$ By representing t as a fraction with coprime integers s and r. You should get that 2 is the only possible prime divisor that cancels. Gerhard "That And Some Algebra Too" Paseman, 2018.01.31. $\endgroup$– Gerhard PasemanCommented Jan 31, 2018 at 21:13
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$\begingroup$ I do not understand how the problem converts to the above problem. By letting t=s/r we get x^2(s^4+r^4)=t^4+s^2r^2. then? $\endgroup$– mehdi baghalaghdamCommented Jan 31, 2018 at 21:23
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$\begingroup$ please explain more about the first case.of course I know that Fermat case has not solution... And what do you think about the second example? $\endgroup$– mehdi baghalaghdamCommented Jan 31, 2018 at 21:30
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1 Answer
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Of course, t=1 is a solution, but we count that as trivial and move on.
As polynomials in t, we have a ratio of (some products of) cyclotomic polynomials, and the gcd for most choices will be $1+ t^d$ when $d$ is gcd of m and n and both m/d and n/d are odd. Removing that as a common factor, it is unlikely that what remains has any chance of yielding a comparable power of a rational number.
For m=2 and n=4, we consider t in reduced form as r/s, and look for coprime integers r and s such that the ratio of $(r^2 + s^2)s^2$ to $r^4 + s^4$ is the ratio of two integer squares. Since r and s are coprime, the only common divisor of the two terms can be 1 or 2. Fermat says no for 1, and Max says no for 2.
For m=3 and n=5, we can divide out by (1+t) and note that the two polynomials are coprime. Again picking r and s coprime, we need $u=r^2 -rs + s^2$ to be a square, and the denominator $D$ to be a square. We then find $D-u^2$ is $rs(u-rs)$ with $u-rs$ also being a square. We also have $D-r^2s^2$ is $r ^2 + s^2$ times the square $u$. While these conditions do not lead to a contradiction, I strongly suspect a similar manipulation will.
Granville recently proved a result regarding certain sequences having primitive prime factors to an odd power. I suspect that can be used to tackle the case n-m is even. I will include the reference later in a comment.
Gerhard "Sorry, The Contradiction Went Elsewhere" Paseman, 2018.02.01.
answered Feb 1, 2018 at 22:51
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$\begingroup$ $x = 1$ instead of $t = 1$, I guess. $\endgroup$– LSpiceCommented Feb 1, 2018 at 23:32
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$\begingroup$ Actually "gives" instead of "is". But I will leave it as is. Gerhard "Solutions Can Also Be Indirect" Paseman, 2018.02.01. $\endgroup$ Commented Feb 1, 2018 at 23:52
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$\begingroup$ thank you very much for your valuable comments. I think that there exist integers n,m such that the equation has nontrivial solutions. What do you think about this? $\endgroup$ Commented Feb 2, 2018 at 16:07
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$\begingroup$ Sorry, I don't know what I was thinking. Clearly $t = 1$ does indeed give a solution. $\endgroup$– LSpiceCommented Feb 2, 2018 at 16:48
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$\begingroup$ thanks, but t=0,1,-1 are trivial solution. $\endgroup$ Commented Feb 2, 2018 at 18:33