How complicated are the roots?

We are given that $N(N + 1)(N + 2) = 5 \times 6 \times 7 = 210 \Longrightarrow N^{3} + 3N^{2} + 2N - 210 = 0$ .

By observation of the original equation $N = 5$ is a solution, so factoring out this root we find that

$N^{3} + 3N^{2} + 2N - 210 = (N - 5)(N^{2} + 8N + 42) = 0$ .

Now $N^{2} + 8N + 42$ does not have real roots as the discriminant is less than $0$ , but we cannot assume than $N$ is real, so we must consider all complex roots as well Now the roots of this quadratic are

$N = \dfrac{-8 \pm \sqrt{64 - 4 \times 42}}{2} = -4 \pm i\sqrt{26}$ .

Next, note that $(N + 6)(N + 7)(N + 8) = N^{3} + 21N^{2} + 146N + 336 =$

$(N^{3} + 3N^{2} + 2N) + 18N^{2} + 144N + 336 = 210 + 18N(N + 8) + 336 = 546 + 18N(N + 8).$

For our complex roots we have that $N(N + 8) = (-4 \pm i\sqrt{26})(4 \pm i\sqrt{26}) = -16 - 26 = -42$ ,

in which case $(N + 6)(N + 7)(N + 8) = 546 + 18 \times (-42) = -210 \ne 11 \times 12 \times 13$ .

So the answer is $\boxed{\text{No}}$ .

N (N+1)(N+2)-5 x 6 x 7=0 and (N+6)(N+7)(N+8)-11 x 12 x 13=0 Both must have same roots(all three roots same) for the condition to apply in general. But the sum of roots is different in both equations. (-3 in equation 1 and -21 in equation 2). This implies that both equations have atleast one different root. Hence, false.(wherever I have mentioned root, I mean complex root.)

Since equation $N(N+1)(N+2) = 5 \times 6 \times 7$ has degree 3, we know that it must have three complex solutions (counting multiplicities). Assume the three solutions were also solutions of the equation $(N+6)(N+7)(N+8) = 11 \times 12 \times 13$ . Subtrating one equation from the other, we obtain an equation of second order, since the leading terms $N^3$ cancel out. Thus, the maximum number of solutions that satisfy both equations is 2. (Note: this is not a contradiction because the first equation could have multiple roots.)

We already know one solution to both equations: 5. Since the difference of the equations results in a polynomial equation with integer coefficients and 5 is one of the solutions, the other solution must also be an integer. (This is true in this case: see comment below, sorry!)

This shows that simultaneous solutions to both equations are integers. But the first solution is easily shown to have no integer solutions other than 5 (with multiplicity one).

This is pretty clear if you think about it with polar coordinates.

Assume without loss of generality that $0 < arg(N) < \pi$ .

Then if $N(N+1)(N+2)$ is real and positive, then $arg(N) + arg(N+1) + arg(N+2) = 2\pi$ . Clearly such a solution exists, where the three arguments are roundabout $\frac{2\pi}{3}$ .

Adding a positive real $x$ to each expression reduces each argument term, so the resulting angles sum to less than $2\pi$ , and so the product $(N+x)(N+1+x)(N+2+x)$ cannot also be a positive real for any $x > 0$ .

Let $P(x) = x(x+1)(x+2)$ . If all the numbers $N$ that satisfied the first equation also satisfied the second one, we would have $P(x) - 5 \times 6 \times 7 = P(x + 6) - 11 \times 12 \times 13$ (Polynomial equality). Now, take the derivative twice and we get that $P''(x) = P''(x+6)$ . But $P''(x)$ is a linear polynomial of the form $ax + b$ so this equation says:

$ax + b = a(x+6) + b \implies ax + b = ax + b + 6a \implies 6a =0 \implies a=0$ But $a \neq 0$ because otherwise the first degree polynomial would be constant.

if u watch first equation only origin is shifted by 6 units ....we will compare constant term obtained after shifting in 1 st euation which will 90 but accroding to given equation it will come 1380... so its false