The only ones of height<=5 are x^5 - 5 x^2 - 3, 2*x^5 - 5*x + 4, and 4*x^5 + 5*x + 4 (modulo x->1/x). The only real root of the 2nd one is (1/(2^(2/5) 5^( 3/5)))(-(150 - 100 Sqrt[2] + Sqrt[10 (4250 - 2999 Sqrt[2])])^( 1/5) + (-150 + 100 Sqrt[2] + Sqrt[10 (4250 - 2999 Sqrt[2])])^( 1/5) - (150 + 100 Sqrt[2] - Sqrt[10 (4250 + 2999 Sqrt[2])])^( 1/5) - (150 + 100 Sqrt[2] + Sqrt[10 (4250 + 2999 Sqrt[2])])^(1/5)) This seems almost a case of "*a stupid man asking such a question to which one hundred wise men would not be able to answer*." Various sources, e.g., http://en.wikipedia.org/wiki/Quintic_function give a pair of equations which have rational roots iff x^5+ax+b is solvable and irreducible, but sometimes reducible cases that fail this test, e.g., {3/8 - (5 x)/8 + x^5, 8/3 + (5 x)/3 + x^5}, yield to the quintic solver, producing obscure radical identities of the form biquadratic = sum of four fifth roots of biquadratics, which paralyze FullSimplify. I don't know if it's proven or merely conjectured that there are no solvable irreducible septic trinomials. See http://www.math.harvard.edu/~elkies/trinomial.html --rwg