Suppose that $$P$$ is the polynomial of least degree with integer coefficients such that $$P(\sqrt{7}+\sqrt{5})=2(\sqrt{7}-\sqrt{5})$$. Find $$P(2)$$.

Let $$x = \sqrt7 + \sqrt5$$. We want a polynomial with integer coefficients such that

$$P(x)=2\bigl(\sqrt7-\sqrt5\bigr).$$

The field $$\mathbb{Q}(\sqrt7,\sqrt5)$$ has basis $$\{1,\sqrt7,\sqrt5,\sqrt{35}\}$$ over $$\mathbb{Q}$$. Powers of $$x$$ expressed in this basis are:

$$x^0 = 1$$

$$x^1 = \sqrt7 + \sqrt5$$

$$x^2 = (\sqrt7+\sqrt5)^2 = 7 + 5 + 2\sqrt{35} = 12 + 2\sqrt{35}$$

Using $$\sqrt7\,\sqrt{35}=7\sqrt5$$ and $$\sqrt5\,\sqrt{35}=5\sqrt7$$,

$$x^3 = x\cdot x^2 = (\sqrt7+\sqrt5)(12+2\sqrt{35}) = 12(\sqrt7+\sqrt5)+2\sqrt{35}(\sqrt7+\sqrt5) = 12\sqrt7+12\sqrt5+14\sqrt5+10\sqrt7 = 22\sqrt7+26\sqrt5.$$

Hence, in the basis $$\{1,\sqrt7,\sqrt5,\sqrt{35}\}$$,

1 → $$(1,0,0,0)$$
$$x$$ → $$(0,1,1,0)$$
$$x^2$$ → $$(12,0,0,2)$$
$$x^3$$ → $$(0,22,26,0).$$

We seek integers $$c_0,c_1,c_2,c_3$$ such that

$$c_0 + c_1x + c_2x^2 + c_3x^3 = 2\sqrt7 - 2\sqrt5.$$

Comparing coefficients in the basis gives the system

Constant term: $$c_0 + 12c_2 = 0$$

$$\sqrt{35}$$ term: $$2c_2 = 0 \;\Longrightarrow\; c_2 = 0,\; c_0 = 0$$

$$\sqrt7$$ term: $$c_1 + 22c_3 = 2$$

$$\sqrt5$$ term: $$c_1 + 26c_3 = -2$$

Subtracting the last two equations: $$4c_3 = -4 \;\Rightarrow\; c_3 = -1.$$ Substituting back: $$c_1 + 22(-1)=2 \;\Rightarrow\; c_1 = 24.$

Thus the desired polynomial of least degree is

$$P(t)=c_0 + c_1 t + c_2 t^2 + c_3 t^3 = 24t - t^3.$$

Indeed, $$P(x)=24x - x^3 = 24($$\sqrt$$7+$$\sqrt$$5)-(22$$\sqrt$$7+26$$\sqrt$$5)=2$$\sqrt$$7-2$$\sqrt$$5,$$ as required. A quadratic polynomial cannot work because in $$1,x,x^2$$ the coefficients of $$$$\sqrt$$7$$ and $$$$\sqrt$$5$$ are always equal, so degree $$3$$ is minimal.

Finally, evaluate at $$t=2$$:

$$P(2) = 24$$\cdot$$2 - 2^3 = 48 - 8 = 40.$$

Answer: 40