Prove $A=\sqrt{x^{2026}-x^{406}+2017}$ is not an integer

Question

I have some difficulty with this problem

prove $A=\sqrt{x^{2026}-x^{406}+2017}$ is not an integer

I'vs tried to prove $x^{2026}-x^{406}+{2017}$ is not a square number by congruent (with 2 -> 4 -> 8 and stuck or 3 -> 9) but seems that it dont help.

Can you guys have me this excercise, please. Thank you

Answer 1

Let's look at $\operatorname{mod} 5$.

All the squares $\operatorname{mod} 5$ are $0, 1, 4$.

Let's iterate on remainder of $x$ $\operatorname{mod} 5$.

Also, keep in mind $2017 \operatorname{mod} 5 = 2$ and $2 \operatorname{mod} 5$ can not be a square.

Case $1$: $x \operatorname{mod} 5 = 0$

Then number $\operatorname{mod} 5$ is $2017$. So this can't be square.

Case $2$: $x \operatorname{mod} 5 = 1$

Then number $\operatorname{mod} 5$ equals $1 - 1 + 2017$, again not square.

Case $3$: $x \operatorname{mod} 5 = 2$

Then number $\operatorname{mod} 5$ equals $$2 ^ {2026} - 2 ^ {406} + 2017 = 4 ^ {1013} - 4 ^ {203} + 2017$$ $$= (-1) - (-1) + 2017$$ again not square.

Case $4$: $x \operatorname{mod} 5 = 3$

Then number $\operatorname{mod} 5$ equals $$3 ^ {2026} - 3 ^ {406} + 2017 = 9 ^ {1013} - 9 ^ {203} + 2017$$ $$= (-1) - (-1) + 2017$$ again not square.

Case $5$: $x \operatorname{mod} 5 = 4$

Then number $\operatorname{mod} 5$ equals $$4 ^ {2026} - 4 ^ {406} + 2017 = (-1) ^ {2026} - (-1) ^ {406} + 2017$$ $$= (1) - (1) + 2017$$ again not square.

Answer 2

By taking only last digit of number x

$$x^{2026}\equiv x^{406}(\bmod 10)$$ $$x^{2026} - x^{406}+2017 \equiv 7(\bmod 10)$$

There's no square of a number ending with $7$.