Piero della Francesca 1412 1492 wrote the following in his

Piero della Francesca (1412 - 1492) wrote the following in his Treatise on Calculation: When things and squares and cubes and squares are equal to a number, one should divide the number of things by the number of cubes, square the result and add to the number. Then the thing will be equal to the square root of the square root of the sum minus the root of the result of dividing things by cubes.

a. Translate this solution in modern notation. Try this out on x^4+2x^3+3x^2+2x=8. Does it work?

b. By working backwards from his solution, can you find a fourth-degree equation type for which his solution will work?

Solution

a) the first sentence obviously refers to a quartic equation A*X^4 + B*X^3 + C*X^2 + D*X = E. The word \"thing\" refers to the modern term \"unknown X\". The second sentence would therefore refer to the quantity (D/B)^2 + E.

For simplicity (and without loss of generality), let us take A = 1. The last sentence is to be interpreted to mean

X = the fourth-root of (D/B)^2 + E minus the square root of D/B
,
which is supposed to give a root of the equation (but not all four roots in those days mathematicians were not yet aware that there should be four roots including complex roots).

However, this is not always true unless it happens that C = 6 times D/B.

apply this to the equation

X^4 + 2*X^3 + 3*X^2 + 2*X = 8

where B = 2, C = 3, D = 2, E = 8, then D/B = 1 and (D/B)^2 + E = 9. Note that this time 6 times D/B = 6, which is NOT equal to 3. If we take X to be the fourth root of 9 minus the square root of 1, we do NOT obtain a root of the equation.

so this does not work

b)\"The equation

X^4 + B*x^3 + C*x^2 + D*x = E

with 8*C = 3*B^2 and 16*D = B^3 has

X = the fourth-root of (D/B)^2 + E minus the square root of D/B

as a root.\"

we can have the eqaution

X^4 + 12*X^3 + 54*X^2 + 108* X = 175