Tessellate S.T.E.M.S. (2019) - Mathematics - Category A (School) - Set 2 - Subjective Problem 2

Find all positive integers n n for which 2n−12^n-1 has a multiple of the form m2+9m^2+9.

This problem is a part of Tessellate S.T.E.M.S. (2019)

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Note by Tessellate S.T.E.M.S. Mathematics
2 years, 7 months ago

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We will show that 2n−12^n-1 has a multiple of the form m2+9m^2+9 if and only if nn is a power of 2.2.

First, suppose there is an odd prime p∣n.p|n. Look at the equation m2≡−9(mod2p−1).m^2 \equiv -9 \pmod{2^p-1}. Since pp is odd, 2p−12^p-1 is not divisible by 3,3, so we can multiply both sides by the multiplicative inverse of 99 to get k2≡−1(mod2p−1).k^2 \equiv -1 \pmod{2^p-1}. The Jacobi symbol (−12p−1)\left( \frac{-1}{2^p-1} \right) is −1-1 because 2p−1≡3(mod4),2^p-1 \equiv 3 \pmod 4, so −1-1 is not a square mod 2p−12^p-1 and the equation has no solutions.

Hence 2p−12^p-1 has no multiple of the form m2+9.m^2+9. Since any multiple of 2n−12^n-1 is a multiple of 2p−1,2^p-1, 2n−12^n-1 has no multiples of the form m2+9.m^2+9.

It remains to show that if nn is a power of 2,2, 2n−12^n-1 has a multiple of the form m2+9.m^2+9. Proceed by induction. First, n=1,2,4n=1,2,4 are clear as base cases. (Use m=6.m=6.) Now suppose nn is a (larger) power of 2.2. By the inductive hypothesis, 2n/2−12^{n/2}-1 has a multiple of the form m12+9.m_1^2+9. Also, (2n/2+1)⋅9=(3⋅2n/4)2+9,(2^{n/2}+1) \cdot 9 = (3 \cdot 2^{n/4})^2+9, so (3⋅2n/4)2+9(3 \cdot 2^{n/4})^2+9 is a multiple of 2n/2+12^{n/2}+1. Now use the Chinese Remainder Theorem to construct a number mm such that m≡m1(mod2n/2−1)m≡3⋅2n/4(mod2n/2+1) \begin{aligned} m &\equiv m_1 \pmod{2^{n/2}-1} \\ m &\equiv 3 \cdot 2^{n/4} \pmod{2^{n/2}+1} \\ \end{aligned}

Then m2+9m^2+9 will be divisible by the relatively prime positive integers 2n/2−12^{n/2}-1 and 2n/2+1,2^{n/2}+1, hence by their product 2n−1.2^n-1. □\Box

(To give an idea of that inductive process, let's construct an mm for n=8.n=8. First, we have that 62+96^2+9 is a multiple of 24−1,2^4-1, and 122+912^2+9 equals 99 times 24+1,2^4+1, so now use the Chinese Remainder Theorem to get an mm satisfying m≡6(mod15)m \equiv 6 \pmod{15}, m≡12(mod17).m \equiv 12 \pmod{17}. One solution, m=−39,m=-39, leads to the multiple 392+9=153039^2+9 = 1530 of the number 28−1=255.2^8-1 = 255.)

Patrick Corn - 2 years, 5 months ago
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