# Integer Root (Tomato Subjective 43)

Show that the equation $$x^3 + 7x – 14(n^2 +1) = 0$$ has no integral root for any integer n.

Solution:

We note that $$14(n^2 +1) – 7x = x^3$$ implies $$x^3$$ is divisible by 7. This implies x is divisible by 7 (as 7 is a prime number). Suppose x= 7x’. Hence we can rewrite the given equation as:

$$7^3 x’^3 + 7 \times 7 x’ – 14 (n^2 +1 ) = 0$$.

Cancelling out a 7 we have $$7^2 {x’}^3 + 7{x’} = 2(n^2 +1)$$. Since 7 divides left hand side, it must also divide the right hand side. Since 7 cannot divide 2, it must divide $$n^2 + 1$$ as 7 and 2 are coprime.  Note that 7 cannot divide $$n^2 +1$$ as square of a number always gives remainder 0, 1, 4, 2 when divided by 7 and never 6. But if $$n^2 + 1$$ is divisible by 7 then $$n^2$$ must give remainder 6 when divided by 7.  Hence contradiction.

Necessary Lemma: square of a number always gives remainder 0, 1, 4, 2 when divided by 7

$$n \equiv 0, \pm 1 , \pm 2 , \pm 3 \mod 7\Rightarrow n^2 \equiv 0, 1, 4, 9 (=2) \mod 7$$

Key Ideas: Modular Arithmetic