Prove that there are no intergers x >1, y > 1, z > 1 with x! + y! = z! Solution

Question

Prove that there are no intergers x >1, y > 1, z > 1 with x! + y! = z!

Explanation / Answer

Let x, y, and z be integers greater than 1. Now let y = x + pand z = x + q where p and q are positive integers (possibly zero ifany of the integers are the same). We may assume x =(3)(4)..(x+p) = j where j > 1, which gives us 1 = j, but j >1, so this is false. Case III Suppose p > q, then we would have 1 = (x+1)(x+2)...(x+q) -(x+1)(x+2)...(x+p) 1 = (x+1)(x+2)...(x+q)[1 - (x+q+1)(x+q+2)...(x+p)] But since x > 1, x is at least 2, so (x+q+1)(x+q+2)...(x+p)>= (3)(4)...(x+p), so: 1 = (x+1)(x+2)...(x+q)k for k < 0. x is greater than 1, so(x+1)(x+2)...(x+q) = j where j > 0, so we get 1 = jk with j > 0 and k < 0, which means 1 = a negativenumber, which is false. Therefore, x! + y! = z! results in a contradiction for any positiveintegers x, y, and z, so it must be false that x! + y! = z!

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.