VI OM - I - Problem 9

Present the polynomial $ x^4 + x^3 + x^2 + x + 1 $ as the difference of squares of two polynomials of different degrees with real coefficients.
If the polynomial $ W (x) = x^4 + x^3 + x^2 + x + 1 $ is equal to the difference $ U(x)^2 - V(x)^2 $, where $ U(x) $ and $ V(x) $ are polynomials of different degrees, then the polynomial $ U(x) $ must be of the second degree and the polynomial $ V(x) $ must be of the first or zero degree. In this case, the polynomial $ V(x)^2 $ does not contain terms of degree higher than two, so the first two terms of the polynomial $ U(x)^2 $ must be the same as in the given polynomial $ W(x) $, i.e., $ U(x)^2 $ has the form $ x^4 + x^3 + \ldots $. From this, we conclude that the polynomial $ U(x) $ has the form $ U(x) = x^2 + \frac{1}{2}x + a $. Therefore,






From the expression obtained for $ V(x)^2 $, we see that $ V(x) $ cannot be of zero degree, as it cannot be simultaneously $ 2a - \frac{3}{4} = 0 $ and $ a - 1 = 0 $; $ V(x) $ must therefore be of the first degree. It follows further that the square of the function $ V(x) $ must be a quadratic trinomial, in which the coefficient of $ x^2 $ is positive, and the discriminant equals $ 0 $, thus






The second condition gives, after simplification, the equation






This equation has three roots $ a = 1 $ and $ a = \frac{1}{4} $ ($ -1 \pm \sqrt{5} $), of which only the root $ a = 1 $ satisfies the condition $ 2a - \frac{3}{4} > 0 $. Ultimately, we obtain the factorization






as the only solution to the problem.