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- Apr 13, 2013

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I am looking at the following exercise:

"Prove that $\forall n \in \mathbb{N}$ the number $17 \cdot 3^{2n+1}+41 \cdot n^2 \cdot 3^{n+1}+2$ " is not a square of an integer."

We do it like that:

$a(n)= 17 \cdot 3^{2n+1}+41 \cdot n^2 \cdot 3^{n+1}+2=3 \cdot (17 \cdot 3^{2n}+41 \cdot n^2 \cdot 3^{n})+2$.

We suppose that $a(n)=b^2$,for a $b\in \mathbb{Z}$

Applying the Euclidean Division,$b$ divided by $3$ gives us $b=3k+r, r\in \{0,1,2\}$.

So,$b^2=(3k+r)^{2}$

For $r=0 , b^2=9k^2=3 \cdot 3k^2$,so the remainder is equal to $0$.

For $r=1 , b^2=9k^2+6k+1=3(3k^2+2k)+1 $,so the remainder is equal to $1$.

For $r=2 , b^2=9k^2+12k+4=3(3k^2+4k+1)+1 $,so the remainder is equal to $1$.

So,the remainder of $b^2$ divided by $3$,cannot be $2$ at any case..So,it cannot be $a(n)=b^2$.

But...Why do we apply the Euclidean Division,$b$ divided by $3$ ??