Find $\lambda\in \mathbb{N}$ such that $4x^{11+\lambda}+7y^{19+\lambda}=1,~5x^6+9y^{14}=1$ has solutions.

Question

Find all positive integers $\lambda$ such that the system:

$$\left \{\begin {array}{lll} 4x^{11+\lambda}+7y^{19+\lambda}=1\\ 5x^6+9y^{14}=1\\ \end{array} \right.$$ has solutions.

I tried to check if some values of $\lambda$ (eg $1,2,3$) give us solutions, but the non-linearity of the system makes me think this is not a good way to start. Any help is appreciated.

Thanks in advance.

Im not sure, but is very likely that the number of solutions is small, so perhaps what you have to do is to prove that $\lambda=1,2,3$ works (and maybe other small values) and then try to prove that for $\lambda>n$ this has no solutions by contradiction. Another option is that you have to distinguish between even and odd $\lambda$, so if I where you I would do the cases $\lambda=1,\dots, 10$ and then try to find a general pattern. — Marcos, Feb 18 '22 at 11:09
Write $x^6=z$ and $y^{14}=w$ and rewrite the equations. I suppose $x,y\in \Bbb C$? — Dietrich Burde, Feb 18 '22 at 11:16
After some simulations with Geogebra, I strongly doubt there are any solutions... — Jean Marie, Feb 18 '22 at 11:37
It has been given as an exercise in Numerical Linear Algebra. — Nikolaos Skout, Feb 18 '22 at 12:52
Sorry but there are $n$ books with this title. Can you give the authors ? — Jean Marie, Feb 18 '22 at 13:03
I find solutions with some negative $\lambda$, e.g., $\lambda=-8$ or $\lambda=-10$. — Jean Marie, Feb 18 '22 at 13:12
@Bazyli Zuczek No for $\lambda=-5$ you do not have real solutions. — Jean Marie, Feb 18 '22 at 16:17
Indeed, as mentioned before, it seems that there are no solutions for positive integer $\lambda$. The graphics seem to be a good approach (though we don't have a formal proof). — Nikolaos Skout, Feb 19 '22 at 11:51
For the real case, it suffices to notice that the second equation implies $|x| < 1 \land |y| < 1$ and therefore $|4 x^{\lambda + 11}| \leq 5 x^6 \land |7 y^{\lambda + 19}| \leq 9 y^{14}$, with one of the inequalities being strict. — Maxim, Feb 19 '22 at 13:51
Please answer my question about the names of the authors of your book. There are at least 6 books with this title. — Jean Marie, Feb 19 '22 at 23:29
Jean Marie, it is given as an exercise in a course called: Numerical Linear Algebra. The source of the exercise is not given (if any). — Nikolaos Skout, Feb 20 '22 at 00:00
@NikolaosSkout how is this related to numerical linear algebra? — cineel, Feb 21 '22 at 18:13

Piquito · Answer 1 · 2022-02-18T20:08:53.677

COMMENT.- All curve $\Gamma$ of equation $ax^{2m}+by^{2n}=1$ where $a,b,m,n$ are positive integers is symmetric to both axis because if $(x,y)$ is a point of $\Gamma$ so is for the four points $(\pm x,\pm y)$. Besides $\Gamma$ has a shape like a square and is contained in the square centered at the origine and having side equal to $2$.

When $m$ and $n$ are increased "in uniform way" (like as in this problem) we can see that the curves are disjoint:

in fact, for example, in the system $$\left \{\begin {array}{lll} 4x^{12}+7y^{20}=1\\ 5x^6+9y^{14}=1\\\end{array} \right.$$ we have in the first equation $x=0\Rightarrow y=\sqrt[20]{\frac14}$ and in the second one $x=0\Rightarrow y=\sqrt[14]{\frac15}$ and $$\sqrt[14]{\frac15}\lt\sqrt[20]{\frac14}\iff0.8914....\lt0.9330...$$

Similarly we can see that making $y=0$ in both equations we have values for $x$ such that the corresponding to the first equation is less than the other.

It clearly follows that there are not solutions for $\lambda$ odd.

(For $\lambda$ even the corresponding curves are not closed but we can reasoning, I guess,in akin way).

The shape of curves $\Gamma$ looks more like a rectangle than a square (I wanted to say this). — Piquito, Feb 18 '22 at 23:54
@Jean Marie: Good, however the case $\lambda$ even is similar to solve with my way. Thank you. — Piquito, Feb 21 '22 at 14:52

Find $\lambda\in \mathbb{N}$ such that $4x^{11+\lambda}+7y^{19+\lambda}=1,~5x^6+9y^{14}=1$ has solutions.

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