Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.

Clearly we have to show that $7$ and $8$ both will divide $n$.

I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd perfect square. - so we have $4n+1\equiv 1\pmod{8}$; from this $2|n$ so $3n+1$ is a odd perfect square. - so $3n+1\equiv 1\pmod{8}$ so $8|n$ but I can't show $7|n$. How do I show this?

Thanks for the help.


5 Answers 5


$3n+1=a^2$ and $4n+1=b^2$ gives $4a^2-3b^2=1$. With $x=2a$ and $y=b$ we get an instance of Pell's equation.


Trying small integers, it is obvious that the minimal solution is $4-3=1$, that is $x_1=2$ and $y_1=1$.

Therefore, all the solutions are of the form $(x_k,y_k)$ with $x_k+y_k\sqrt 3=(x_1+y_1\sqrt 3)^k$ or equivalently $$ \begin{align} x_{k+1}&=2x_k+3y_k\\ y_{k+1}&=x_k+2y_k, \end{align} $$ (which already works starting from the trivial solution $x_0=1$, $y_0=0$).

This gives $3y_{k+1} =3x_k+6y_k$ which implies $$\begin{align}&x_{k+2}-2x_{k+1}=3x_k+2x_{k+1}-4x_k\\ \Leftrightarrow &x_{k+2}-4x_{k+1}+x_k=0 \end{align}$$ and $x_0=1$, $x_1=2$.

Clearly, the $x_i$ alternate in sign and we are only interested in the even ones which are those with odd index. $$\begin{align}x_{2k+1} &= 4x_{2k}-x_{2k-1}\\&=4(4x_{2k-1}-x_{2k-2})-x_{2k-1}\\ &=15x_{2k-1}-x_{2k-1}-x_{2k-3}\\&=14x_{2k-1}-x_{2k-3}.\end{align}$$

Upon division by two, we obtain the recurrence for the solutions $a_k$ to $3n+1=a^2$ subject to the other condition.

$$a_k=14a_{k-1}-a_{k-2},$$ with $a_0=1$ and $a_1=13(=\dfrac{x_3}2)$.

Now, it is trivial to check that modulo $3$, all $a_k$ are $\equiv 1$ because $14\cdot 1 - 1 \equiv 1 \mod 3$. Which implies that $3 \mid a^2-1$.

Furthermore, modulo $7$, all $a_k$ are $\equiv \pm 1$ because $a_k \equiv - a_{k-2} \mod 7$ which implies that $7 \mid a^2-1$.

And finally, modulo $4$, all $a_k$ are $\equiv 1$ because $14-1 \equiv 1 \mod 4$ which implies that $8 \mid (a-1)(a+1)=a^2-1$.

So, in summary, $56 \mid \dfrac{a^2-1}3$ as desired.


$3n+1 = a^2$ and $4n+1 = b^2$. As $b^2 \equiv 1 \mod 4 \implies b \equiv \pm 1 \mod 4$. So, $4n +1 = (4k_1 \pm 1)^2 \implies n = 2k_1 (2k_1 \pm 1)$. So, $2 \mid n$ , let $n = 2n'$. $3n +1 = 2(3n')+1$ hence $a$ is odd, say $a = 2a'+1$. $6n' +1 = 4a'^2 + 4a' +1 $ $\implies 3n' = 2a'(a'+1) $. $2 \mid a'(a'+1) \implies 4 \mid n'$ (as $4 \nmid 3$). So $\boxed{8 \mid n}$

$x^2 \equiv 0, 1,2,-3 \mod 7$. If $3n+1 \equiv 0 \implies n \equiv 2 \implies 4n+1 \equiv 2$ (All in mod 7). If $3n+1 \equiv 1 \implies n \equiv 0 \implies 4n+1 \equiv 1$. If $3n+1 \equiv 2 \implies n \equiv -2 \implies 4n+1 \equiv 0$. If $3n+1 \equiv -3 \implies n \equiv 1 \implies 4n+1 \equiv -2$ (not possible). So, $n \equiv 0,2,-2 \mod 7$.

Now, we will show that $n \equiv 2 \mod 7$ is not possible. As $n \equiv 0 \mod 8 \implies n \equiv 16 \mod 56$. Let $n = 56n'' +16$. $a^2 = 3(56n'' +16)+1 = 168n'' +49$. So $7 \mid a^2 \implies 7 \mid a \implies 49 \mid a^2$. But $49 \nmid 168$ hence a contradiction. Similarly we can show that $n \equiv -2 \mod 7$ is ruled out. So $\boxed{7 \mid n}$

Hence $\boxed{56 \mid n}$ as $gcd(8,7) = 1$. $\Box$


$3n+1=x^2$ $4n+1=y^2$ $4x^2-3y^2=1$ Put $z=2x\ \ \ $ $z^2-3y^2=1$

One of solution is: $(z_0,y_0)=(2,1) $

Other solutions are given by: $(z_k+ \sqrt3 y_k)=(2+ \sqrt 3)^{k} $

From above we get, $z_{k+1}=7z_k+12y_k $ And $y_{k+1}=4z_k+7y_nk $ (we use $(2+\sqrt3)^2$ to maintain parity of $z$)

Or (in terms of $x_k $);

$x_{k+1}=7x_k+6y_k\ \ \ \ $ $y_{n+1}=8x_k+7y_k $


$x_{k+1} \equiv -y_k \pmod 7$

$y_{k+1} \equiv x_k \pmod 7$

Since $y_0^2 - x_0^2 \equiv 0 \pmod 7$, the claim follows by induction.


Let $3n+1=K^2$ and $4n+1=P^2$, where $P$ and $K$ are integers. Therefore, we know that $4(K^2)-3(P^2)=1$. Rearranging terms gives us $K^2= \frac {1+3P^2}4$. By testing values and guesswork we know that $P=13$ and $K=15$. Therefore $3n+1=169$, and then solving for $n$ gives us $n=56$.


For $7 | n$, consider adding $4n+1$ and $3n+1$.

Observe that we get $7n + 2 = x^2 + y^2$, where we can let $4n+1 = x^2$, and $3n+1 = y^2$. Now, we see that mod $7$, the only quadratic residues are $0$, $1$, $2$, $4$. The only way to get two squares summing to 2 mod 7 is either if both are congruent to $1$ mod $7$, or if one is $0$ mod $7$ and one is $2$ mod $7$.

If both are congruent to $1$ mod $7$, we have n is $0$ mod $7$, so we need to show that the other case is not possible.

We now try to multiply $4n+1$ by $3n+1$, to get $12n^2 + 7n + 1$. This mod 7 is $5n^2 + 1$. If one of the squares if $0$ mod 7, it means that the product is also 0 mod 7, meaning that $5n^2 + 1 \equiv 0 $ mod $7$. This means that $5n^2 \equiv 6 $ mod $7$, and hence that $n^2 \equiv 5 $ mod $7$, which is an impossibility due to quadratic residues mod 7.


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