Certain amount of numbers t=x^3+y^2 in a given set

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Certain amount of numbers t=x^3+y^2 in a given set

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Source: VJIMC 2017, Category II, Problem 4

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#1 h Apr 2, 2017, 3:45 PM • 3 Y

Y by shinichiman, anantmudgal09, Adventure10

A positive integer $t$ is called a Jane's integer if $t = x^3+y^2$ for some positive integers $x$ and $y$ . Prove
that for every integer $n \ge 2$ there exist infinitely many positive integers $m$ such that the set of $n^2$ consecutive
integers $\{m+1,m+2,\dotsc,m+n^2\}$ contains exactly $n + 1$ Jane's integers.

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#2 h Apr 3, 2017, 11:48 AM • 4 Y

Y by anantmudgal09, Tawan, Adventure10, Mango247

Nice problem.

This post has been edited 1 time. Last edited by shinichiman, Apr 3, 2017, 1:27 PM
Reason: typo

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#3 h Apr 20, 2017, 7:17 PM • 3 Y

Y by shinichiman, Tawan, Adventure10

Note that Claim 1. in the above solution can be done in an easier way;

Suppose there exist only finite $m$ that $\{ m+1, \ldots, m+n^2 \}$ contains at most $n$ Jane's integers.
There exist constant $C$ that if $m>C$ , then the number of Jane\s integers in such set is more than $n$ .
Since the number of Jane's integer in the interval $[1,N^6]$ is at most $N^{\frac{6}{3}}\times N^{\frac{6}{2}}=N^5$
Consider large enough $N^6=C+Tn^2$ where $T$ is a positive number, we get that the number of Jane's integers not larger than $N^6$ is at most $N^5$ but also at least $0+Tn=\frac{N^6-C}{n}$ .
So $N^5\geq \frac{N^6-C}{n}$ for all positive integer $N^6>C$ and fixed positive constant $C,n$ .
This gives a contradiction for large enough $N$ , done.

This post has been edited 2 times. Last edited by ThE-dArK-lOrD, Apr 20, 2017, 7:19 PM

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#4 h Apr 20, 2017, 7:31 PM • 4 Y

Y by shinichiman, Tawan, Adventure10, Mango247

Maybe the most difficult part when solving this problem is really to note that the request for exactly $n+1$ such numbers is a bit of a red herring.
Another problem where this "continuity principle" is essential is this problem (which is somewhat easy once you got that idea, but otherwise quite hard to solve...)

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#5 h Apr 23, 2017, 9:45 AM • 3 Y

Y by Tawan, Adventure10, Mango247

My solution is almost identical to the ones posted here. But the idea is too nice for me to resist posting

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#6 h Mar 13, 2025, 11:25 PM

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If I am not mistaken, for $n\geq 3$ and any $x$ one has $f(x^6) \geq n+1$ ? The integers $(x^2)^3 + j^2, j=1,\ldots,n$ , as well as $x^6 + 8 = 2^3 + (x^3)^2$ , are good. However, there does not seem to be an easy example for $n=2$ and the Pell argument is essential there.

This post has been edited 1 time. Last edited by Assassino9931, Mar 13, 2025, 11:28 PM

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