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58 Chapter 2. On a conjecture about addition modulo 2 k − 12.5.4 A polynomial expressionWe can now prove a first step towards Proposition 2.5.1. We show that SX d is a product ofexponentials in basis 2 and 4 (but not only 4 yet!) by multivariate polynomials.{ }} { { }} { { }} {Proposition 2.5.14. For j 2 > 0 and X = ( 0, . . . , 0, 1, . . . , 1, 2, . . . , 2),whereΞ d X =j∑2−1i=02 −i Σ d−j2+1+iX∑ j 0= 4− i=1 βi j∑2−13 j1l=0S d X = 2j23 j2 d∏( j2−1∑i=l2 −ii!i=j 0+j 1+1j 0[ ] )i ∑lj 1((1 − 4 −βi ) S d−j2k+k j0 +1+...+k j0 +j 1 =lXj 2− Ξ d X),() ( )j kl∑ 0β i Π d X .k, k j0+1, . . . , k j0+j 1Ξ d X is a sum for I ⊂ {j 0 + 1, . . . , j 0 + j 1 } of terms of the form 4 − ∑ j 0i=1 βi−∑ i∈I βi multiplied bya multivariate polynomial of degree in β i exactly j 2 if i ∈ I, j 2 − 1 if 1 ≤ i ≤ j 0 , 0 otherwise, andof total degree j 2 + #I − 1.Proof. The proof goes by induction on j 2 ≥ 1.For j 2 = 1, this is Proposition 2.5.7.Suppose now that j 2 > 1. From Proposition 2.5.7,and by induction hypothesis on j 2 ,S d X = 2 1 − 4−β d3= 2j23 j2 d∏i=j 0+j 1+1Using Proposition 2.5.8, we haveso thatwhence the proposition.S d X = 2j23 j2 d∏SX d = 2 1 − 4−β dS d−1X3− 2T X d ;2 j2−1 d−1∏3 j2−1i=j 0+j 1+1((1 − 4 −βi ) S d−j2T d X = 13 j2 d∏i=j 0+j 1+1= 2j23 j2 d∏i=j 0+j 1+1((1 − 4 −βi ) S d−j2Xi=j 0+j 1+1((1 − 4 −βi ) S d−j2X)− Ξ d−1X− 2TX d .(1 − 4 −βi )Σ d X ,X((1 − 4 −βi ) S d−j2X− Ξ d−1X− Ξ d X))− Ξ d−1X− 2TXd− 2−j2+1 Σ d X,)i=1