Maths Olympiad Prep

Library / /6 of 19

Number theory Difficulty 5.5 AIME, harder Prove it North Macedonia

Find all the pairs (m,n)(m,n) of integers which satisfy the equation
m5n5=16mn.m^5 - n^5 = 16mn.

Solution

If one of mm, nn is 00, the other has to be 00 too, and (m,n)=(0,0)(m,n) = (0,0) is one solution.

If mn0mn \neq 0, let d=gcd(m,n)d = \gcd(m,n) and we write m=dam = da, n=dbn = db, a,bZa, b \in \mathbb{Z} with (a,b)=1(a,b) = 1. Then, the given equation is transformed into
d3a5d3b5=16ab(1) d^3 a^5 - d^3 b^5 = 16ab \quad (1)
So, by the above equation, we conclude that ad3b5a | d^3 b^5 and thus ad3a | d^3. Similarly bd3b | d^3. Since (a,b)=1(a,b) = 1, we get that abd3ab | d^3, so we can write d3=abrd^3 = abr with rZr \in \mathbb{Z}. Then, equation (1) becomes
abr5abr3=16abr(a5b5)=16 abr^5 - abr^3 = 16ab \Rightarrow r(a^5 - b^5) = 16
Therefore, the difference a5b5a^5 - b^5 must divide 1616. This means that
a5b5=±1,±2,±4,±8,±16. a^5 - b^5 = \pm 1, \pm 2, \pm 4, \pm 8, \pm 16.
The smaller values of a5b5|a^5 - b^5| are 11 or 22. Indeed, if a5b5=1|a^5 - b^5| = 1 then a=±1a = \pm 1 and b=0b = 0 or a=0a = 0 and b=±1b = \pm 1, a contradiction. If a5b5=2|a^5 - b^5| = 2, then a=1a = 1 and b=1b = -1 or a=1a = -1 and b=1b = 1. Then r=8r = -8, and d3=8d^3 = -8 or d=2d = -2. Therefore, (m,n)=(2,2)(m,n) = (-2,2).

If a5b5>2|a^5 - b^5| > 2 then, without loss of generality, let a>ba > b and a2a \ge 2. Putting a=x+1a = x+1 with x1x \ge 1, we have
a5b5=(x+1)5b5(x+1)5x5==5x4+10x3+10x2+5x+131 \begin{aligned} |a^5 - b^5| &= |(x+1)^5 - b^5| \ge |(x+1)^5 - x^5| = \\ &= |5x^4 + 10x^3 + 10x^2 + 5x + 1| \ge 31 \end{aligned}
which is impossible. Thus, the only solutions are (m,n)=(0,0)(m,n) = (0,0) or (2,2)(-2,2).

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.