Algebra_IMO

Solution 1. Call a number q good if every number in the se ond line appears in the third lineun onditionally. We rst show that the numbers 0 and ˘2 are good. The third line ne essarily ontains 0, so 0 is good. For any two numbers a, b in the rst line, write a “ x´y and b “ u´v,where x, y, u, v are (not ne essarily distin t) numbers on the napkin. We may now writ2ab “ 2px ´ yqpu ´ vq “ px ´ vq 2 py ´ uq 2 ´ px ´ uq 2 ´ py ´ vq 2 ,whi h shows that 2 is good. By negating both sides of the above equation, we also see that ´2is good.We now show that ´2, 0, and 2 are the only good numbers. Assume for sake of ontradi tionthat q is a good number, where q R t´2, 0, 2u. We now onsider some parti ular hoi es of numbers on Gugu's napkin to arrive at a ontradi tion.Assume that the napkin ontains the integers 1, 2, . . . , 10. Then, the rst line ontainsthe integers ´9, ´8, . . . , 9. The se ond line then ontains q and 81q, so the third line mustalso ontain both of them. But the third line only ontains integers, so q must be an integer.Furthermore, the third line ontains no number greater than 162 “ 9 2 9 2 ´ 0 2 ´ 0 2 or lessthan ´162, so we must have ´162 ď 81q ď 162. This shows that the only possibilities for q are ˘1.Now assume that q “ ˘1. Let the napkin ontain 0, 1, 4, 8, 12, 16, 20, 24, 28, 32. The rstline ontains ˘1 and ˘4, so the se ond line ontains ˘4. However, for every number a in therst line, a ı 2 pmod 4q, so we may on lude that a 2 ” 0, 1 pmod 8q. Consequently, everynumber in the third line must be ongruent to ´2, ´1, 0, 1, 2 pmod 8q; in parti ular, ˘4 annotbe in the third line, whi h is a ontradi tion.Solution 2. Let q be a good number, as dened in the rst solution, and dene the polynomialPpx1, . . . , x10q asźiăjpxi ´ xj q źaiPS qpx1 ´ x2qpx3 ´ x4q ´ pa1 ´ a2q 2 ´ pa3 ´ a4q 2 pa5 ´ a6q 2 pa7 ´ a8q 2 ˘ ,where S “ tx1, . . . , x10u.We laim that Ppx1, . . . , x10q “ 0 for every hoi e of real numbers px1, . . . , x10q. If any twoof the xi are equal, then Ppx1, . . . , x10q “ 0 trivially. If no two are equal, assume that Guguhas those ten numbers x1, . . . , x10 on his napkin. Then, the number qpx1 ´ x2qpx3 ´ x4q is inthe se ond line, so we must have some a1, . . . , a8 so thatqpx1 ´ x2qpx3 ´ x4q ´ pa1 ´ a2q 2 ´ pa3 ´ a4q 2 pa5 ´ a6q 2 pa7 ´ a8q 2 “ 0,Shortlisted problems  solutions 15and hen e Ppx1, . . . , x10q “ 0.Sin e every polynomial that evaluates to zero everywhere is the zero polynomial, and theprodu t of two nonzero polynomials is ne essarily nonzero, we may dene F su h thatFpx1, . . . , x10q ” qpx1 ´ x2qpx3 ´ x4q ´ pa1 ´ a2q 2 ´ pa3 ´ a4q 2 pa5 ´ a6q 2 ` pa7 ´ a8q 2 ” 0 (1)for some parti ular hoi e ai P S .Ea h of the sets ta1, a2u, ta3, a4u, ta5, a6u, and ta7, a8u is equal to at most one of the foursets tx1, x3u, tx2, x3u, tx1, x4u, and tx2, x4u. Thus, without loss of generality, we may assumethat at most one of the sets ta1, a2u, ta3, a4u, ta5, a6u, and ta7, a8u is equal to tx1, x3u. Letu1, u3, u5, u7 be the indi ator fun tions for this equality of sets: that is, ui “ 1 if and only if tai , aii 1u “ tx1, x3u. By assumption, at least three of the ui are equal to 0.We now ompute the oe ient of x1x3 in F . It is equal to q q 2pu1 u3 ´ u5 ´ u7q “ 0,and sin e at least three of the ui are zero, we must have that q P t´2, 0, 2u, as desired.