A nested square root that looks like Pell's equation but needs less machinery

CAPTAIN_TOAD·Jun 18, 2026

Writing

m^2-12n^2=1

suggests parametrizing Pell solutions. Is there a way to prove the inner half-integer is a square without classifying all those solutions? Problem: For a positive integer

n

it is known that the number

\sqrt{12n^2 + 1}

also is a positive integer. Prove that the number

\sqrt{\frac{\sqrt{12n^2 + 1} + 1}{2}}

also is a positive integer. Source: Baltic Way 2019 · MNET-01G1

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3 replies

shulk_said_no·Jun 19, 2026

m is odd, so write m=2k+1. The equation becomes

k(k+1)=3n^2

. Consecutive integers are coprime, so their prime exponents force either k=3a², k+1=b², or k=a², k+1=3b².

CAPTAIN_TOAD·Jun 19, 2026

The second possibility would give a²≡−1≡2 modulo 3, impossible. Therefore k+1 is a square, which is exactly the quantity inside the outer root.

savefile_07·Jun 19, 2026

This avoids solving Pell entirely. The crucial point is that all primes other than 3 occur to even exponent in the product, and coprimality assigns each prime to only one of the consecutive factors.

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