Both \(4n+1\) and \(3n+1\) are perfect squares.

Then prove that \(56|n\).

From "An Excursion in Mathematics" . (Solution is not given/Challenge Problem)

Pls type your solution in comments section.

(I could prove easily that \(8|n\) and \(n=7k\ or\ 7k-2 \ or\ 7k+2\))

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TopNewestIf \( 4n+1 = a^2 \) and \( 3n+1 = b^2 \) then \( (2b)^2-3a^2 = 1 \). The solutions to Pell's equation are well known, and \( b \) is even in only every other one. Basically you look at powers of the "fundamental unit" \( 2 + \sqrt{3} \) and set \( 2b \) equal to the first summand and \( a \sqrt{3} \) equal to the second one. For the first term to be even, you need the odd powers of \( 2 + \sqrt{3} \) only.

This leads to the recurrence relation \( a_1 = b_1 = 1 \), \( a_{k+1} = 7a_k + 8b_k, b_{n+1} = 6a_k + 7b_k \). After that it's an easy induction to see that \( a_k \) and \( b_k \) are always \( \pm 1 \) mod \( 7 \), so \( n = (b^2-1)/3 \) should be divisible by \( 7 \).

This is by no means the easiest or best solution, but I thought I should mention that one can generate all the values of \( n \) satisfying the equations: \( 0, 56, 10920, \ldots \). – Patrick Corn · 3 years ago

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– Megh Parikh · 3 years ago

Problem solution by basic method.I don't know Pellsequation :) Thanks for the helpLog in to reply

My half solution:

\(4n+1=a^2\) is \(odd\).

Therefore \(n=2k\) as any perfect square is of form \(8k +1\ OR\ 4k\)

Then \(3n+1\) is also odd.

therefore \(3n+1=8k+1 \implies n=8k\)

for 7 see remainders mod 7 and \[a^2+b^2\equiv2\ (mod\ 7) \] then see \[a^2-b^2\] – Megh Parikh · 3 years ago

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