**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**

ISI B. Math solved questions needed