## Problems Problem 1 Solve in integers the equation \( x^2+y^2-1=4xy \). Substituting \( u=y-2x \) reduces the equation to \( u^2=3x^2+1 \) whose general solution is given by \( x+u\sqrt3=(2+\sqrt3)^n \). Problem 2
For a given integer \( d \), solve \( x^2-dy^2=1 \) in the set of No knowledge of Pell’s equation is required here. For \( x\neq 1 \) we have \( d\left(\frac y{x-1}\right)=\frac{x+1}y \). Setting \( \frac y{x-1}=t\in\mathbb{Q} \) we easily obtain \( x=\frac{dt^2+1}{dt^2-1} \) and \( y=\frac{2t}{dt^2-1} \). Problem 3 Let \( (x,y)=(a,b) \), \( a,b\in\mathbb{N} \) be the smallest integer solution of \( x^2-dy^2=1 \). Consider the sequence defined by \( y_0=0 \), \( y_1=b \), \( y_{n+1}=2ay_n-y_{n-1} \) for \( n\geq1 \). Show that \( ay_n^2+1 \) is a square for each \( n \). Show that if \( ay^2+1 \) is a square for some \( y\in\mathbb{N} \), then \( y=y_n \) for some \( n \). Let \( x_n+y_n^{\prime}=\left(a+b\sqrt d\right)^n \). All solutions \( (x,y) \) of \( x^2=ay^2+1 \) are given by \( (x,y)=(x_n,y_n^{\prime}) \) for some \( n \). We have \[ y_n^{\prime}=\frac1{2\sqrt d}\left(\left(a+b\sqrt d\right)^n-\left(a-b\sqrt d\right)^n\right).\quad\quad\quad\quad\quad (1)\] Since \( a\pm b\sqrt d \) are the solutions of the quadratic equation \( x^2-2ax+1=0 \), relation (1) easily implies that \( y_{n+2}^{\prime}- 2ay_{n+1}^{\prime}+y_n=0 \). Therefore the sequences \( (y_n) \) and \( (y_n^{\prime}) \) satisfy the same initial conditions and recurrent relation and must be equal. The statement of the problem follows immediately. Problem 4 Prove that \( 5x^2+4 \) or \( 5x^2-4 \) is a perfect square if and only if \( x \) is a term in the Fibonacci sequence. Problem 5 Find all \( n\in\mathbb{N} \) such that \( \displaystyle\binom n{k-1}= 2\binom nk+\binom n{k+1} \) for some natural number \( k< n \). Multiplication by \( \frac{(k+1)!(n-k+1)!}{n!} \) transforms the equation into \( k(k+1)=2(k+1)(n-k+1)+(n-k)(n-k+1) \), which is simplified as \( n^2+3n+2=2k^2+2k \), i.e. \[ (2n+3)^2+1= 2(2k+1)^2.\] The smallest solution of equation \( x^2-2y^2=-1 \) is \( (1,1) \), and therefore all its solutions \( (x_i,y_i) \) are given by \( x_i+y_i\sqrt2=(1+\sqrt2)^{2i+1} \). Note that \( x_i \) and \( y_i \) are always odd, so \( n=\frac{x_i-3}2 \) is an integer and a solution to the problem. Clearly, there are no other solutions. Problem 6 Let \( a\in\mathbb{N} \) and \( d=a^2-1 \). If \( x,y \) are integers and the absolute value of \( m=x^2-dy^2 \) is less than \( 2a+2 \), prove that \( |m| \) is a perfect square. The smallest solution of \( N(z)=1 \) is \( z_0=a+\sqrt d \). If \( N(z)=m \) has a solution, then by the theorem \ref{T.5} it also has a solution \( z=x+y\sqrt d \) in which \( x\leq\frac{z_0+1}{2\sqrt{z_0}}\sqrt{|m|}= \sqrt{\frac{a+1}2|m|} \). For \( |m|< 2a+2 \) this inequality becomes \( x< a+1 \), and thus \( (x,y)=(a,1) \) and \( m=1 \) or \( y=0 \) and \( m=x^2 \). Problem 7 Prove that if \( m=2+2\sqrt{28n^2+1} \) is an integer for some \( n\in\mathbb{N} \), then \( m \) is a perfect square. We start by finding those \( n \) for which \( m \) is an integer. The pair \( (\frac m2-1,n) \) must be a solution of Pell’s equation \( x^2-28y^2=1 \) whose smallest solution is \( (x_0,y_0)= (127,24) \); hence \( \frac m2-1+n\sqrt{28}=(127+24\sqrt{28})^k \) for some \( k\in\mathbb{N} \). Now we have \[ m=2+(127+24\sqrt{28})^k+(127-24\sqrt{28})^k=A^2,\] where \( A=(8+3\sqrt7)^k+(8-3\sqrt7)^k \) is an integer. Problem 8 Prove that if the difference of two consecutive cubes is \( n^2 \), \( n\in\mathbb{N} \), then \( 2n-1 \) is a square. Let \( (m+1)^3-m^3=3m^2+3m+1=n^2 \). Then \( (2n)^2= 3(2m+1)^2+1 \), so \( (2n,2m+1) \) is a solution of Pell’s equation \( x^2-3y^2=1 \). As shown already, we obtain \( 2n+(2m+1)\sqrt3= (2+\sqrt3)^l \). In order for \( n \) to be integral, \( l \) must be odd. It follows that \( 4n=(2+\sqrt3)^{2k+1}+(2-\sqrt3)^{2k+1} \). Finally, \[ 2n-1=\frac{(1+\sqrt3)^2(2+\sqrt3)^{2k}+(1-\sqrt3)^2 (2-\sqrt3)^{2k}-8}4=N^2,\] whereby \( N \) is an integer: \( \displaystyle N=\frac12\left((1+\sqrt3)(2+\sqrt3)^k+(1-\sqrt3) (2-\sqrt3)^k\right) \). Problem 9 If \( n \) is an integer such that \( 3n+1 \) and \( 4n+1 \) are both squares, prove that \( n \) is a multiple of 56. Let us find all such \( n \). Denoting \( 3n+1=a^2 \) and \( 4n+1=b^2 \) we have \( (2a)^2-3b^2=1 \). We shall find all solutions of \( x^2-3b^2=1 \) with an even \( x=2a \). The solutions of the Pell’s equation \( u^2-3v^2=1 \) are given by \( (u,v)=(u_k,v_k) \), where \( u_k+v_k\sqrt3=(2+\sqrt3)^k \). One easily sees that \( u_k \) is even if and only if \( k \) is odd. Thus \( (x,b)= (u_{2k+1},v_{2k+1}) \), where we also have \[ 2u_{2k+1}=(2+\sqrt3)^{2k+1}+(2-\sqrt3)^{2k+1}.\] It follows that \( (a,b)=(\frac12u_{2k+1},v_{2k+1}) \) and \( n=\frac13(a^2-1)=\frac1{12}(u_{2k+1}^2-4) \), which yields \[ \begin{array}{rcl}48n&=&\displaystyle(7+4\sqrt3)^{2k+1}+(7-4\sqrt3)^{2k+1}-14\\ &=&\displaystyle 2\left(7^{2k+1}-7+48\binom{2k+1}2 7^{2k-1}+48^2\binom{2k+1}2 7^{2k-3}+\cdots\right).\end{array}\] We thus obtain a simpler expression for \( n \), making the divisibility by \( 56 \) obvious: \[ n=\frac{7^{2k+1}-7}{24}+2\binom{2k+1}2 7^{2k-1}+2\cdot48 \binom{2k+1}27^{2k-3}+\cdots\] Problem 10 Prove that the equation \( x^2-dy^2=-1 \) is solvable in integers if and only if so is \( x^2-dy^2=-4 \). Problem 11 Let \( p \) be a prime. Prove that the equation \( x^2-py^2=-1 \) has integral solutions if and only if \( p=2 \) or \( p\equiv 1 \) (mod 4). If the considered equation has a solution \( (x,y) \), then \( p\mid x^2+1 \); hence either \( p=2 \) or \( p\equiv 1 \) (mod 4). For \( p=2 \), \( x=y=1 \) is a solution. We shall show that there is a solution for each prime \( p=4t+1 \). A natural starting point (and the only one we see!) is the existence of an integral solution \( (x_0,y_0) \) with \( x_0,y_0\in\mathbb{N} \) to the corresponding Pell’s equation: \( x_0^2-py_0^2=1 \). We observe that \( x_0 \) is odd: otherwise \( y_0^2\equiv py_0^2\equiv 3 \) (mod 4). Thus in the relation \( x_0^2-1=(x_0-1)(x_0+1)=py_0^2 \), factors \( x_0+1 \) and \( x_0-1 \) have the greatest common divisor 2, and consequently one of them is a doubled square (to be denoted by \( 2x^2 \)) and the other one \( 2p \) times a square (to be denoted by \( 2py^2 \)). The case \( x_0+1=2x^2 \), \( x_0-1=2py^2 \) is impossible because it leads to a smaller solution of Pell’s equation: \( x^2-py^2=1 \). It follows that \( x_0-1=2x^2 \), \( x_0+1=2py^2 \) and therefore \( x^2-py^2=-1 \). Problem 12 If \( p \) is a prime of the form \( 4k+3 \), show that exactly one of the equations \( x^2-py^2=\pm2 \) has an integral solution. This is very similar to the previous problem. At most one has a solution. Now if \( (x_0,y_0) \) is the smallest solution of the corresponding Pell’s equation, if \( x_0\pm1 \) are not coprime, the equality \( (x_0-1)(x_0+1)=py_0^2 \) gives us \( x_0\pm1=2x^2 \) and \( x_0\mp1=2py^2 \), i.e. \( x^2-py^2=\pm1 \), which is impossible in either case. Therefore \( x_0\pm1 \) are coprime, which implies \( x_0\pm1=x^2 \), \( x_0\mp1=py^2 \) and \( x^2-py^2=\pm2 \). Problem 13 Prove that \( 3^n-2 \) is a square only for \( n=1 \) and \( n=3 \). Problem 14 Prove that if \( \displaystyle \frac{x^2+1}{y^2}+4 \) is a perfect square, then this square equals 9. |

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