Two Polynomials

OOPS! [2] I did not see this!

f(x)=a_1\left(x-\alpha \right)(x-\beta)(x-\gamma)(x+\alpha+\beta+\gamma).

g(x)=a_2\left(x-p \right)(x-q)(x-r)(x+p+q+r)

From the fact that f(\alpha)=f(\beta)=f(\gamma)=f(\alpha+\beta+\gamma)=0, we arrive at the conclusion that the roots of g(x) and f(x) are same.

Now we need to evaluate l and m where it is given that [ln]=[mn]

Where n=\prod({x-\alpha})....Needless to mention that l and m are just the constants.

WE see that l=m is the only solution in this case.
Proving this is no big deal if l and m are rationals. However if they are irrationals then one can say,

[l]=x_1x_2x_3...x_n & \left\{l \right\}=y_1y_2y_3...y_m... - i.e let it reccur.

Similarly let [m]=x'_1x'_2...x'_k & \left\{m \right\}=y'_1y'_2...y'_s...

I chhose n=10^t, where t is any arbitary large natural.

That itself completes the story proving \boxed{f(x)=g(x)} \forall x belonging to reals. [4]

Its easily proved that f(x) and g(x) need to have the same sign for their leading coefficients (otherwise for large enough x, f and g will have opposite signs)

WLOG let the leading coefficients be +ve.

Then both f,g→∞ as x→∞

That means f and g take on integer values at infinitely many x.

i.e. h(x) = f(x) - g(x) =0 for infinitely many x.

But h(x) is of degree at most 4.

Hence h(x)≡0 i.e. f(x)≡g(x)