# Number theory problem

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 $$n=7k\ or\ 7k-2 \ or\ 7k+2$$)

Note by Megh Parikh
4 years, 5 months ago

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## Comments

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If $$4n+1 = a^2$$ 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: $$0, 56, 10920, \ldots$$.

- 4 years, 5 months ago

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Problem solution by basic method.I don't know Pellsequation :) Thanks for the help

- 4 years, 5 months ago

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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 $a^2-b^2$

- 4 years, 5 months ago

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