I have following equation: $$n^2 - n + 2\equiv 0\pmod{49}$$
So I get: $$n^2 - n + 2\equiv 0\pmod{7}$$ The only number is: $$n\equiv 4 \pmod7$$
Thus, I used Hensel's Lemma. And according to (3) point (http://math453spring2009.wikidot.com/hensel-s-lemma) We see that there is no sulution when we have $7^2$.
What about my solution ? Is it correct ? Maybe somebody has other idea?
On
More generally, if $$ x^2 - x y + 2 y^2 \equiv 0 \pmod {49}, $$ then both $x,y \equiv 0 \pmod 7.$
BECAUSE $2$ is invertible $\bmod 7,$ and $$ 4 x^2 - 4 x y + 8 y^2 = (2x-y)^2 + 7 y^2 \equiv 0 \pmod {49}, $$ so first $$ (2x-y) \equiv 0 \pmod 7. $$ However, this gives $$ (2x-y)^2 \equiv 0 \pmod {49}, $$ with $$ 7 y^2 \equiv 0 \pmod {49}, $$ $$ y^2 \equiv 0 \pmod 7,$$ $$ y \equiv 0 \pmod 7. $$ Finally $$ 2x \equiv 0 \pmod 7 $$ and $$ x \equiv 0 \pmod 7 $$
So, there is no solution with my letter $y=1,$ meaning the original question has no integer solutions.
A solution without Hansel's Lemma: suppose there is an integer $n$ such that $n^2-n+2$ is divisible by $49$. Then $$ n^2-n+2=n^2-8n+16+7n-14=(n-4)^2+7n-14 $$ is divisible by $7$ so $(n-4)^2$ is divisible by $7$, which further implies $n-4$ is divisible by $7$. So you actually have $(n-4)^2$ is divisible by $49$. Now, write $n=7m+4$. Observe that $$ 7n-14=7(7m+4)-14=49m+14, $$ which isn't divisible by $49$. Contradiction.