Diophantine equation $x^p+ax=y^p+by$

Question

Problem. Is there a prime number $p$ (desirably $p\le 3$) and an infinite set $A\subset\mathbb N$ such that for any distinct numbers $a,b\in A$ the Diophantine equation $x^p+ax=y^p+by$ has no positive integer solutions?

Answer 1

My suspicion is that every prime except $2$ has this property. I gave up on $p=2$ after realizing that (except for a=1,2,4) $x^2+ax=y^2$ always has solutions.

Here is some circumstantial evidence for $p=3,$ which should be harder than larger $p.$

If my calculations are correct, the following set $$A=\{0,40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 62, 64, 65, 66,$$ $$ 67, 68, 69, 70, 72, 73, 74, 75, 76, 78, 80, 81, 82, 84, 85, 88, 92, 93, 94, 96, 100, 102\}$$ of size $44$ is such that there are no solutions of $x^3+ax=y^3+by$ in positive integers $y \gt x$ with $a \gt b$ and $a,b\in A.$

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Here are some details:

It is pretty fast to check if $x^3+ax=y^3+by$ has integer solutions. We may assume $a \gt b \geq 0.$ If any solutions occur, they must have $ \frac{a-b-3}3 \gt x:$

To have $x^3+ax=y^3+by$ requires that $y \geq x+1.$ Thus there can only be solutions provided that $$x^3+ax \geq (x+1)^3+b(x+1) \gt x^3+3x^2+3x+bx$$ So $$(a-b-3)x>3x^2.$$

Actually, I don't think that $x$ can be anywhere near that large. Here are all the cases $[a,[x,y]]$ with $0 \lt x \lt y\ $ and $x^3+ax=y^3$ such that $ a \leq 104:$

$ [7, [1, 2]], [26, [1, 3]], [28, [2, 4]], \mathbf{[38, [4, 6]], [61, [8, 10]]}, [63, [1, 4], [3, 6]], [104, [2, 6]]$

So the other $97$ values of $a$ with $1 \leq a \leq 103$ are all compatible with being in $A$ if $0 \in A.$ But some pairs are not compatible with each other. Reducing to the set $A$ listed above makes all pairs compatible.

It is a fairly ad-hoc set selected by making a graph with the $97$ values as vertices and all $323$ bad pairs $(a,b)$ as edges, deleting some vertices of highest degree and repeating.

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Note that whenever $[a,[x,y]]$ makes $x^3+ax=y^3$, so also does $[d^2a,[dx,dy]].$ Also $[a,[x,y]]=[k^3-1,[1,k]]$ makes $x^3+ax=y^3.$ These two facts together account for five of the seven pairs above with $a \lt 104,$ the other two are in bold.

Up to $a=5000$ there are only $140$ values of $a$ incompatible with $0 \in A.$ $114$ of them have only one bad pair $x,y$ with $x^3+ax=y^3$ and the other $26$ have two bad pairs. About half of the bad values of $a$ are of the form $a=d^2(k^3-1)$ and explained by the two facts above.

Of these $140$ there are $46,28,25,23$ and $18$ respectively in the intervals $[1,1000],[1001,2000],[2001,3000],[3001,4000]$ and $[4001,500].$

Answer 2

There are such sets with arbitrarily large length for any fixed $p\ge3$.

Proof:

Let $$x^p+ax=y^p+by.$$ Then $$x^p-y^p=by-ax.$$ Suppose $a\gt b$. Then $y\lt x$, so $y\le x-1$. Since $y\le x-1$, $$x^p-y^p\ge x^p-(x-1)^p$$ and $$by-ax\le b(x-1)-ax=(b-a)x-b$$, giving $$x^p-(x-1)^p\le (b-a)x-b,$$ or $$x^p-(x-1)^p\le tx-b,$$ where $t=b-a$, giving $b=a+t$. Since the LHS is greater than $$px^{p-1}$$ for all positive integer $x$ (which can be proved by using the binomial theorem), this implies that $$px^{p-1}\lt tx-b,$$ so one can take $t$ smaller than $c\cdot b^{\frac{1}{p-1}}$, where the constant $c$ depends only on $p$.

Hence the cardinality of the set $A$ can be arbitrarily large, in fact $$A=\{a,a+1,a+2,...,a+t\}.$$