${\left\{\alpha_{n+2}=2 \alpha_{n+1}-\alpha_{n}+2,\alpha_{0}=0\right\}}$ ${\left\{\beta_{n+2}=2 \beta_{n+1}-\beta_{n},\beta_{0}=8\right\}}$ We can solve it easily, since it's linear ${\alpha_{n}=n^{2}+c_{1}n-n}$ ${\beta_{n}=n c_{2}+8}$ ${\alpha_{n}^{2}+\beta_{n}^{2}=\left(n c_{2}+8\right)^{2}+\left(n^{2}+c_{1}n-n\right)^{2}}$ Seting ${n=1}$ Making a Pythagorean Triple ${c_{1}=a^{2}-b^{2}}$, ${c_{2}+8=2 a b}$ ${c_{1}=(a-b) (a+b)}$ ,${c_{2}=2 (a b-4)}$ Substituing on the previosly equation ${\alpha_{n}=n \left(a^{2}-b^{2}+n-1\right)}$ ${\beta_{n}=2 (a b-4) n+8}$ $a,b$ are just integers numbers. just set $n=1992$ and you have infinitely many solutions

Hello Yagaron, yagaron wrote: ${\left\{\alpha_{n+2}=2 \alpha_{n+1}-\alpha_{n}+2,\alpha_{0}=0\right\}}$ ${\left\{\beta_{n+2}=2 \beta_{n+1}-\beta_{n},\beta_{0}=8\right\}}$ We can solve it easily, since it's linear ${\alpha_{n}=n^{2}+c_{1}n-n}$ ${\beta_{n}=n c_{2}+8}$ I'm ok till this point (although, in a previous post - now deleted -, I said it was wrong ). But I don't agree with the reminder. I rewrite the expressions : ${\alpha_{n}=n^{2}+a n}$ where $a=a_{1}-1$ ${\beta_{n}=b n+8}$ where $b=b_{1}+8$ ${\alpha_{n}^{2}+\beta_{n}^{2}=n^{4}+2an^{3}+(a^{2}+b^{2})n^{2}+16b n+64}$ We can easily identify this with ${\alpha_{n}^{2}+\beta_{n}^{2}=(n^{2}+a n+8)^{2}}$ which gives $a^{2}+b^{2}=a^{2}+16$ and $16 a n=16 b n$ And two immediate solutions : $a=b=4$ and $a=b=-4$ and : $\alpha_{1992}=1992^{2}+4*1992$ and $\beta_{1992}= 4*1992+8$ and $\alpha_{1992}=1992^{2}-4*1992$ and $\beta_{1992}=-4*1992+8$ -- Patrick

i made the following $a_{n}^{2}+b_{n}^{2}$ is a perfect square for every $n \geq 0$ but in the expressions ${\alpha_{n}=n^{2}+c_{1}n-n}$ ${\beta_{n}=n c_{2}+8}$ i have 2 unknow constantes. how to find then!? to simmplify i just set $n=1$ and made a pythagorean triple with the squares of the 2 sequences. and make a substituition for the 2 constantes to a,b. but i forget that this is valid only for $n=1$ i just changed the constants. i did not find then... i forget to review the function. sorry. but i still think that there is more solutions for the constants. c ya

yagaron wrote: ${\left\{\alpha_{n+2}=2 \alpha_{n+1}-\alpha_{n}+2,\alpha_{0}=0\right\}}$ ${\left\{\beta_{n+2}=2 \beta_{n+1}-\beta_{n},\beta_{0}=8\right\}}$ We can solve it easily, since it's linear ${\alpha_{n}=n^{2}+c_{1}n-n}$ ${\beta_{n}=n c_{2}+8}$ ${\alpha_{n}^{2}+\beta_{n}^{2}=\left(n c_{2}+8\right)^{2}+\left(n^{2}+c_{1}n-n\right)^{2}}$ ... just set $n=1992$ and you have infinitely many solutions OK, the first part is right, but not the second one. It is true that $\alpha_n^2+\beta_n^2=n^4+2(c_1-1)n^3+c_2^2n^2+(c_1-1)^2n^2+16c_2 n+64$ must be a perfect square, but for all $n$. Let $k_1>\frac{c_2^2}{2}>k_2$, where $k_1,k_2$ are integers. Then, $\alpha_n^2+\beta_n^2-(n^2+(c_1-1)n+k_1)^2=(c_2^2-2k_1)^2n^2+(16c_2-2k_1(c_1-1))n+64-k_1^2$, where for sufficiently large $n$, the first term dominates and is negative, hence $\alpha_n^2+\beta_n^2<(n^2+(c_1-1)n+k_1)^2$ for infinitely many values of $n$. Similarly, $\alpha_n^2+\beta_n^2>(n^2+(c_1-1)n+k_2)^2$ for infinitely many values of $n$, or $k=\frac{c_2^2}{2}$ must be an integer, and $\alpha_n^2+\beta_n^2=(n^2+(c_1-1)n+k)^2$ for infinitely many values of $n$, or for infinitely many values of $n$, \[0=\alpha_n^2+\beta_n^2-\left(n^2+(c_1-1)n+\frac{c_2^2}{2}\right)^2=c_2(16-c_2(c_1-1))n+\frac{256-c_2^2}{4}\] It suffices with two of them, so that from one side, $c_2^2=256$ for $c_2=\pm16$, yielding $c_1-1=1$ or $c_1-1=-1$, for $c_1=2$ or $c_1=0$ respectively. There are therefore exactly two pairs of values $(\alpha_1,\beta_1)$, and the corresponding values of $(a_{1992},b_{1992})$ can easily be found. The problem might as well have read "find all possible values of..." instead of "find at least two values of...".