Maths Olympiad Prep

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Number theory Difficulty 7.6 National olympiad, round 2 Prove it

A natural number is called [i]bright [/i] if it is the sum of a perfect square and a perfect cube.
Prove that if rr and ss are any two positive integers, then
(a) there exist infinitely many positive integers nn such that both r+nr+n and s+ns+n are [i]bright[/i],
(b) there exist infinitely many positive integers mm such that both rm and sm are [i]bright[/i].

Solution

### Part (a)
1. Let k=rs k = r - s .
2. There exist infinitely many zN z \in \mathbb{N} such that 2z3+k1 2 \nmid z^3 + k - 1 . This means that z3+k1 z^3 + k - 1 is odd.
3. Define (x,y)=(z3+k2,z3+k22) (x, y) = \left( \frac{z^3 + k}{2}, \frac{z^3 + k - 2}{2} \right) . Since z3+k z^3 + k and z3+k2 z^3 + k - 2 are both even, x x and y y are integers.
4. We have:
x2+1=y2+z3+k x^2 + 1 = y^2 + z^3 + k
5. Let n=y2+z3s n = y^2 + z^3 - s . Then:
s+n=y2+z3 s + n = y^2 + z^3
and
r+n=x2+13 r + n = x^2 + 1^3
6. For sufficiently large z z , n>0 n > 0 .

Thus, there exist infinitely many positive integers n n such that both r+n r + n and s+n s + n are bright. \blacksquare

### Part (b)
1. Let k k be an arbitrary positive integer.
2. Define:
m=r2s3+k6s2r3 m = r^2 s^3 + k^6 s^2 r^3
3. Then:
sm=(rs2)2+(k2sr)3 sm = (rs^2)^2 + (k^2 sr)^3
and
rm=(sr2k3)2+(rs)3 rm = (sr^2 k^3)^2 + (rs)^3

Thus, there exist infinitely many positive integers m m such that both rm rm and sm sm are bright. \blacksquare

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.