Off-diagonal Rado number for $x+y+c=z$ and $x+y+k=z$

Off-diagonal Rado n um b er for x + y + c = z and x + y + k = z Ra jat A dak a, ∗ , Y ash Bakshi a , L. Sunil Chandran a , Saraswati Girish Nanoti a a Dep artment of Computer Scienc e and Automation, Indian Institute of Scienc e, Bangalor e, India Abstract The study of Ramsey-t yp e problems for linear equations originated with Sc hur’s theorem and w as later placed in a systematic framework b y Ric hard Rado. In the off-diagonal setting, one fixes a pair of distinct linear equations ( E 1 , E 2 ) and asks for the least in teger N suc h that every red–blue coloring of { 1 , 2 , . . . , N } must yield either a red solution to E 1 or a blue solution to E 2 . This threshold integer is referred to as the off-diagonal Rado num b er of the system ( E 1 , E 2 ) . In this w ork, w e study the discrete and contin uous off-diagonal Rado n umber for non-homogeneous linear system of equations x + y + c = z and x + y + k = z where c ≤ k . W e determine the exact tw o-color discrete and con tin uous off-diagonal Rado n umber R 2 ( c, k ) asso ciated with this system of equations. Keywor ds: Off-diagonal Rado num b er, Contin uous Rado num b er, Sc h ur-type equations 2020 MSC: 05D10, 05C55 1. In tro duction Sc hur’s classical theorem [11] establishes that every finite coloring of the p ositiv e integers N m ust con tain a mono c hromatic solution to the equation x + y = z . Building on this, Rado [8] pro vided a complete c haracterization of those systems of linear equations that are r e gular , meaning that for ev ery finite coloring of the p ositiv e integers, there exists a mono c hromatic solution to the system. While the early literature concentrated largely on diagonal settings, where a single equation must ha v e a monochromatic solution, attention has more recen tly shifted tow ard other v arian ts lik e off-diagonal, disjunctive or simultane ous . In off-diaganol framew ork, rather than requiring a single equation to b e solved in a single color, one seeks to guaran tee that at least one of t wo distinct equations admits a mono c hromatic solution, with each equation asso ciated with a differen t color. This viewp oin t naturally extends classical Sc hur-t yp e problems b y allo wing differen t equa- tions to b e asso ciated with different colors. As a result, it leads to a richer framework that rev eals new combinatorial b eha vior and pro vides deeper structural understanding of the problem. ∗ Corresp onding author Email addr esses: rajatadak@iisc.ac.in (Ra jat Adak), yashbakshi@iisc.ac.in (Y ash Bakshi), sunil@iisc.ac.in (L. Sunil Chandran), saraswatig@iisc.ac.in (Sarasw ati Girish Nanoti) Definition 1.1 (Rado num b er) . Let E b e a linear equation with integer co efficien ts. The t-c olor R ado numb er of E , denoted by R t ( E ) , is the smallest p ositiv e integer N suc h that ev ery coloring using t -colors of the set [1 , N ] contains a mono c hromatic solution to E . If no suc h integer N exists, then R t ( E ) is said to b e infinite. Definition 1.2 (Off-diagonal Rado num ber) . Let E 1 and E 2 b e linear equations with integer co efficien ts. The two-c olor off-diagonal R ado numb er of the system of equations ( E 1 , E 2 ) , denoted by R 2 ( E 1 , E 2 ) , is the smallest p ositiv e integer N suc h that every red–blue coloring of the set [1 , N ] con tains either a red solution to E 1 , or a blue solution to E 2 . If no suc h integer N exists, then R 2 ( E 1 , E 2 ) is said to b e infinite. Definition 1.3 (Off-diagonal con tinuous Rado num ber) . Let E 1 and E 2 b e linear equations with real co efficien ts, and let α ∈ R . The two-c olor off-diagonal c ontinuous R ado numb er of the system of equations ( E 1 , E 2 ) on [ α, ∞ ) , denoted b y R 2 ( E 1 , E 2 ; α ) , is the smallest real n umber N ≥ α suc h that every red–blue coloring of the in terv al [ α , N ] contains either a red solution to E 1 , or a blue solution to E 2 . If no such real num ber N exists, then R 2 ( E 1 , E 2 ; α ) is said to b e infinite. 2. Previous W ork The study of Rado num b ers originates from classical results in Ramsey theory . Sc hur [11] pro ved that the equation x + y = z is regular, in tro ducing what are no w kno wn as Sch ur n umbers. Rado [8] significantly generalized this by characterizing all partition-regular linear systems, thereb y identifying precisely whic h systems admit mono c hromatic solutions under ev ery finite coloring. Burr and Lo o [3] initiated the systematic study of such problems, particularly for non- homogeneous equations like x + y + c = z , providing b ounds and constructions for several families. This line of work was further developed in recent studies [13]. Beutelspac her and Bresto v ansky [2] established a fundamen tal result sho wing that for an y p ositiv e integer k , the 2 -color Rado num b er of x 1 + x 2 + · · · + x k = x 0 is k 2 + k − 1 . A natural extension of this theory is the off-diagonal setting, where instead of a single equation, one considers a pair of equations ( E 1 , E 2 ) and seeks the smallest integer N suc h that every red–blue coloring of [1 , N ] yields either a red solution to E 1 or a blue solution to E 2 . This framew ork generalizes classical Rado num b ers by allo wing different equations to b e asso ciated with differen t colors. The first systematic study in this direction w as carried out b y Rob ertson and Sc haal [9], who introduced off-diagonal generalized Sch ur num b ers. F or p ositiv e integers k and ℓ , they determined the smallest in teger S ( k , ℓ ) such that every red–blue coloring of [1 , S ( k , ℓ )] con- tains either a red solution to x 1 + x 2 + · · · + x k = x 0 2 or a blue solution to x 1 + x 2 + · · · + x ℓ = x 0 . F urther dev elopmen ts were made by My ers and Rob ertson [6], who established general b ounds for homogeneous linear equations ( x + q y = z , x + sy = z ) and obtained exact v alues in several sp ecial cases. Later, Y ao and Xia [14] deriv ed exact form ulas for sp ecific families suc h as R 2 (2 x + q y = 2 z , 2 x + sy = 2 z ) , for o dd in tegers q > s ≥ 1 . In 2007, Sab o and Sc haal [10] studied non-homogeneous equations such as x 1 + x 2 + c = x 3 and determined their disjunctive Rado num b ers. In 2025, V estal [13] inv estigated the same equations in the context of sim ultaneous solutions. More recen tly , an off-diagonal rado n umber for com bination of homogeneous and non-homogeneous equations has b een also studied [1]. In this paper, w e study the off-diagonal Rado num b ers for these equations, thereb y con tinuing the inv estigation of the ma jor Ramsey-t yp e v ariants asso ciated with them. 3. Main Result 3.1. The discr ete version Theorem 1. Let c and k b e p ositiv e in tegers with 1 ≤ c ≤ k . Then the tw o-color off-diagonal Rado num ber R 2 ( c, k ) for the system x + y + c = z (red) , x + y + k = z (blue) , is: R 2 ( c, k ) =      ∞ , if c ≡ k (mod 2) k + 3 c + 5 , if k ≤ 2 c 2 k + c + 4 if k > 2 c Pr o of. W e denote x + y + c = z as the c -equation and x + y + k = z as the k -equation. W e ev aluate R 2 ( c, k ) case by case. Case 1. Both c and k ha ve different parities. First, supp ose c is even, and k is o dd. Color every o dd integer red and ev ery ev en in teger blue. If x, y , z were a red solution to x + y + c = z , then x, y , z would all b e o dd. Ho wev er, the sum of tw o o dd n umbers and an ev en n umber is ev en, so z w ould b e even, a con tradiction. Similarly , if x, y , z were a blue solution to x + y + k = z , then x, y , z would all b e ev en. But the sum of tw o ev en n umbers and an o dd num b er is o dd, so z would b e o dd, again a con tradiction. Hence, this coloring av oids b oth t yp es of mono c hromatic solutions. If c is o dd and k is even, rev erse the coloring: color ev ery even integer red and ev ery o dd in teger blue. In this case, any solution to x + y + c = z w ould require x, y , z to b e even, 3 but the sum of t wo even n umbers and an o dd n umber is o dd; therefore, this is imp ossible. Lik ewise, any blue solution to x + y + k = z would require x, y , z to b e o dd, but the sum of t wo o dd num b ers and an even num b er is ev en, therefore this is again imp ossible. Th us, for ev ery N there exists a coloring of [1 , N ] with no red solution to x + y + c = z and no blue solution to x + y + k = z . Therefore, the off-diagonal Rado n umber is infinite. Case 2. Same parit y of c and k , and k ≤ 2 c . Claim 3.1. R 2 ( c, k ) ≥ k + 3 c + 5 Pr o of. Let R = { 1 , 2 , . . . , c + 1 } ∪ { k + 2 c + 4 , . . . , k + 3 c + 4 } , B = { c + 2 , c + 3 , . . . , k + 2 c + 3 } . Consider the 2-coloring of [1 , k + 3 c + 4] defined by χ ( n ) = ( Red , n ∈ R, Blue , n ∈ B . Under this coloring, we will sho w that there is no r e d solution to c -equation and no blue solution to k -equation with all v ariables in [1 , k + 3 c + 4] . No blue solution to the k -equation: If x and y are blue, then c +2 ≤ x, y ≤ k + 2 c + 3 and, hence z = x + y + k ≥ ( c +2) + ( c + 2)+ k = k + 2 c + 4 which is red so, for any x , y ∈ B , z is either red or out of range. So there is no blue solution to the k -equation. No red solution to the c -equation: If x and y are red from the first part of the interv al, then z will b e blue as z = x + y + c ≥ 1 + 1 + c = c + 2 and z ≤ ( c + 1) + ( c + 1) + c = 3 c + 2 and since c ≤ k , 3 c + 2 < k + 2 c + 3 . If x comes from the first part and y from the second part of the Red in terv al, then z ≥ 1 + ( k + 2 c + 4) + c = k + 3 c + 5 , but z should lie in the in terv al [1 , k + 3 c + 4] . Hence, there is no red solution to the c − equation. Consequen tly , this coloring a voids red solution to the c -equation and blue solution to the k -equation on [1 , k + 3 c + 4] . Therefore, the off-diagonal Rado n umber satisfies R 2 ( c, k ) ≥ k + 3 c + 5 . Claim 3.2. R 2 ( c, k ) ≤ k + 3 c + 5 Pr o of. W e sho w that any red–blue coloring of [1 , k + 3 c + 5] gives either a red solution to x + y + c = z or blue solution to x + y = z . Assume for the sake of contradiction, there is a coloring suc h that no red solution to x + y + c = z and no blue solution to x + y = z exists, resp ectiv ely . W e start with an arbitrary coloring of n umbers starting with 1 (and using n umbers up to k + 3 c + 5 ), and try to extend this coloring suc h that the desired conditions are satisfied, but end up with either a red solution to the first equation or a blue solution to the second equation. 4 Notation. W e use the conv en tion that, when trying to a void mono chromatic (red or blue) solutions, a giv en coloring with a particular solution is represented as an ordered triple ( x, y , z ) —may force an extension of the coloring. F or example if 1 is red, taking x = y = 1 in c -equation, to av oid a red solution, w e get, (1 , 1 , c + 2) c = ⇒ c + 2 is blue . Sub case 2.1. Assume that 1 is colored red. (1 , 1 , c + 2) c = ⇒ c + 2 is blue . ( c + 2 , c + 2 , k + 2 c + 4) k = ⇒ k + 2 c + 4 is red . ( k + c 2 + 2 , k + c 2 + 2 , k + 2 c + 4) c = ⇒ k + c 2 + 2 is blue . (1 , k + c + 3 , k + 2 c + 4) c = ⇒ k + c + 3 is blue .  k + c 2 + 2 , k + c 2 + 2 , 2 k + c + 4  k = ⇒ 2 k + c + 4 is red . (1 , k + 2 c + 4 , k + 3 c + 5) c = ⇒ k + 3 c + 5 is blue . (2 c + 3 , c + 2 , k + 3 c + 5) k = ⇒ 2 c + 3 is red . (1 , 2 c + 3 , 3 c + 4) c = ⇒ 3 c + 4 is blue . (2 c − k + 2 , c + 2 , 3 c + 4) k = ⇒ 2 c − k + 2 is red . (1 , 2 c − k + 2 , 3 c − k + 3) c = ⇒ 3 c − k + 3 is blue . ( k + 2 , k + 2 , 2 k + c + 4) c = ⇒ k + 2 is blue . Th us, ( k + 2 , 3 c − k + 3 , k + 3 c + 5) is a blue solution to the k -equation, a contradiction. Sub case 2.2. 1 is colored blue. (1 , 1 , k + 2) k = ⇒ k + 2 is red . ( k + 2 , k + 2 , 2 k + c + 4) c = ⇒ 2 k + c + 4 is blue .  k + c 2 + 2 , k + c 2 + 2 , 2 k + c + 4  k = ⇒ k + c 2 + 2 is red .  k + c 2 + 2 , k + c 2 + 2 , k + 2 c + 4  c = ⇒ k + 2 c + 4 is blue . ( c + 2 , c + 2 , k + 2 c + 4) k = ⇒ c + 2 is red . ( c + 2 , c + 2 , 3 c + 4) c = ⇒ 3 c + 4 is blue . (1 , 3 c + 4 , k + 3 c + 5) k = ⇒ k + 3 c + 5 is red . ( k + c + 3 , c + 2 , k + 3 c + 5) c = ⇒ k + c + 3 is blue . Th us (1 , k + c + 3 , 2 k + c + 4) k is a blue solution to the k -equation, a con tradiction. Thus, w e arrive at a contradiction in b oth the sub cases. Therefore, R 2 ( c, k ) ≤ k + 3 c + 5 . 5 F rom Claim 3.1 and 3.2 w e get that when c ≤ k < 2 c , R 2 ( c, k ) = k + 3 c + 5 . Remark 3.3. Note that k + 3 c + 5 > 2 k + c + 4 as k ≤ 2 c and 2 c − k + 2 > 0 . Also, k + c 2 is an integer owing to the same parity of c . Case 3. Same parit y of c and k and k > 2 c . Claim 3.4. R 2 ( c, k ) ≥ 2 k + c + 4 . Pr o of. Let R = { k + 2 , . . . , 2 k + c + 3 } , B = { 1 , 2 , . . . , k + 1 } . Consider the 2 -coloring of [1 , 2 k + c + 3] defined by , χ ( n ) = ( Red , n ∈ R, Blue , n ∈ B . W e show that under this coloring, there is neither a red solution to the c -equation, nor a blue solution to the k -equation. No blue solution to the k -equation: If b oth x and y ∈ B , then z = x + y + k ≥ 1 + 1 + k = k + 2 . Th us, z ∈ R and therefore z is red. Therefore, there is no blue solution to the k -equation. No red solution to the c -equation: If b oth x and y ∈ R , then z = x + y + k ≥ ( k + 2) + ( k + 2) + c = 2 k + c + 4 which is out of range. Therefore, no red solution to c -equation exists. W e get R 2 ( c, k ) > 2 k + c + 3 as a low er b ound. Hence, R 2 ( c, k ) ≥ 2 k + c + 4 . Claim 3.5. R 2 ( c, q ) ≤ 2 k + c + 4 Pr o of. Assume for the sak e of con tradiction, there is a coloring of { 1 , 2 , . . . , 2 k + c + 4 } suc h that no red solution to the c -equation and no blue solution to the k -equation exists, resp ectiv ely . W e deriv e a con tradiction by analyzing the color assigned to 1 . Sub case 3.1. 1 is colored red. (1 , 1 , c + 2) c = ⇒ c + 2 is blue . ( c + 2 , c + 2 , k + 2 c + 4) k = ⇒ k + 2 c + 4 is red . (1 , k + c + 3 , k + 2 c + 4) c = ⇒ k + c + 3 is blue . ( k + c 2 + 2 , k + c 2 + 2 , k + 2 c + 4) c = ⇒ k + c 2 + 2 is blue . Supp ose k − 2 c is red, (1 , k − 2 c, k − c + 1) c = ⇒ k − c + 1 is blue . 6 ( k − c + 1 , c + 2 , 2 k + 3) k = ⇒ 2 k + 3 is red . (1 , 2 k + 3 , 2 k + c + 4) c = ⇒ 2 k + c + 4 is blue . Th us ( k + c 2 + 2 , k + c 2 + 2 , 2 k + c + 4) k is a blue solution, a contradiction. Supp ose k − 2 c is blue, ( c + 2 , k − 2 c, 2 k − c + 2) k = ⇒ 2 k − c + 2 is red . ( k − c + , k − c + , 2 k − c + 2) c = ⇒ k − c + is blue . ( k − c + , c + 2 , 2 k + 3) k = ⇒ 2 k + 3 is red . (1 , 2 k + 3 , 2 k + c + 4) c = ⇒ 2 k + c + 4 is blue . Th us ( k + c 2 + 2 , k + c 2 + 2 , 2 k + c + 4) k is a blue solution, a contradiction. Sub case 3.2. 1 is colored blue. (1 , 1 , k + 2) k = ⇒ k + 2 is red. ( k + 2 , k + 2 , 2 k + c + 4) c = ⇒ 2 k + c + 4 is blue. (1 , k + c + 3 , 2 k + c + 4) k = ⇒ k + c + 3 is red. Supp ose k − 2 c is red, ( k − 2 c, k + 2 , 2 k − c + 2) c = ⇒ 2 k − c + 2 is blue. ( k − c 2 + 1 , k − c 2 + 1 , 2 k − c + 2) k = ⇒ k − c 2 + 1 is red. Th us ( k − c 2 + 1 , k − c 2 + 1 , k + 2) c is a red solution, a contradiction. Supp ose k − 2 c is blue, ( k + c 2 + 2 , k + c 2 + 2 , 2 k + c + 4) k = ⇒ k + c 2 + 2 is red. ( k − 2 c, 3 c + 4 , 2 k + c + 4) k = ⇒ 3 c + 4 is red. ( c + 2 , c + 2 , 3 c + 4) c = ⇒ c + 2 is blue. ( c + 2 , c + 2 , k + 2 c + 4) k = ⇒ k + 2 c + 4 is red. Th us ( k + c 2 + 2 , k + c 2 + 2 , k + 2 c + 4) c is a red solution, a contradiction. Thus, w e arrive at a con tradiction in b oth the sub cases. Therefore, R 2 ( c, k ) ≤ 2 k + c + 4 . F rom Claim 3.4 and 3.5 w e get that when k > 2 c , R 2 ( c, k ) = 2 k + c + 4 . Th us, w e get the tw o-color off-diagonal Rado num ber for all three cases. Now w e extend this to a tw o-color contin uous off-diagonal Rado num b er. 7 3.2. The Continuous version Theorem 2. Let c and k b e real num b ers with 1 ≤ c ≤ k . Then the tw o-color off-diagonal con tinuous Rado n umber R 2 ( c, k ) on the interv al [ α , N ] for the system x + y + c = z (red) , x + y + k = z (blue) , is: R 2 ( c, k ) =      ∞ , if c ≡ k ( mo d 2) k + 3 c + 5 α , if k ≤ 2 c 2 k + c + 4 α if k > 2 c Pr o of. Case 1. Both c and k ha ve different parities. It follows from the discrete v ersion that one may color the integers according to parity , assigning one color to the ev en in tegers and the other to the o dd integers, in suc h a wa y that there is no red solution to x + y + c = z and no blue solution to x + y + k = z . Therefore, the off-diagonal Rado num b er is infinite. Case 2. Both c and k ha ve the same parity and k ≤ 2 c . Claim 3.6. R 2 ( c, k ) = k + 3 c + 5 α The low er b ound is giv en by the follo wing coloring. R = [ α, c + α ) ∪ [ k + 2 c + 4 α , k + 3 c + 5 α ) , B = [ c + 2 α, k + 2 c + 3 α ) . Therefore, R 2 ( c, k ) ≥ k + 3 c + 5 α F or the upp er b ound, again we can lo ok at coloring of the interv al ( α , k + 3 c + 5 α ) . Sub case 2.1. Assume that α is colored red. ( α, α, c + 2 α ) c = ⇒ c + 2 α is blue . ( c + 2 α, c + 2 α , k + 2 c + 4 α ) k = ⇒ k + 2 c + 4 α is red . ( k + c 2 + 2 α, k + c 2 + 2 α, k + 2 c + 4 α ) c = ⇒ k + c 2 + 2 α is blue . ( α, k + c + 3 α, k + 2 c + 4 α ) c = ⇒ k + c + 3 α is blue .  k + c 2 + 2 α, k + c 2 + 2 α, 2 k + c + 4 α  k = ⇒ 2 k + c + 4 α is red . ( α, k + 2 c + 4 α, k + 3 c + 5 α ) c = ⇒ k + 3 c + 5 α is blue . (2 c + 3 α, c + 2 α , k + 3 c + 5 α ) k = ⇒ 2 c + 3 α is red . 8 ( α, 2 c + 3 α, 3 c + 4 α ) c = ⇒ 3 c + 4 α is blue . (2 c − k + 2 α , c + 2 α, 3 c + 4 α ) k = ⇒ 2 c − k + 2 α is red . ( α, 2 c − k + 2 α, 3 c − k + 3 α ) c = ⇒ 3 c − k + 3 α is blue . ( k + 2 α, k + 2 α, 2 k + c + 4 α ) c = ⇒ k + 2 α is blue . Th us, ( k + 2 α, 3 c − k + 3 α, k + 3 c + 5 α ) is a blue solution to the k -equation, a con tradiction. Sub case 2.2. α is colored blue. ( α, α, k + 2 α ) k = ⇒ k + 2 α is red . ( k + 2 α, k + 2 α, 2 k + c + 4 α ) c = ⇒ 2 k + c + 4 α is blue .  k + c 2 + 2 α, k + c 2 + 2 α, 2 k + c + 4 α  k = ⇒ k + c 2 + 2 α is red .  k + c 2 + 2 α, k + c 2 + 2 α, k + 2 c + 4 α  c = ⇒ k + 2 c + 4 α is blue . ( c + 2 α, c + 2 α , k + 2 c + 4 α ) k = ⇒ c + 2 α is red . ( c + 2 α, c + 2 α , 3 c + 4 α ) c = ⇒ 3 c + 4 α is blue . ( α, 3 c + 4 α, k + 3 c + 5 α ) k = ⇒ k + 3 c + 5 α is red . ( k + c + 3 α , c + 2 α, k + 3 c + 5 α ) c = ⇒ k + c + 3 α is blue . Th us ( α , k + c + 3 α, 2 k + c + 4 α ) is a blue solution to the k -equation, a con tradiction. Thus, w e arrive at a contradiction in b oth the sub cases. Therefore, R 2 ( c, k ) ≤ k + 3 c + 5 α . Case 3. Both c and k ha ve the same parity and k > 2 c . Claim 3.7. R 2 ( c, k ) = 2 k + c + 4 α The low er b ound is giv en by the follo wing coloring. R = [ k + α , 2 k + c + 4 α ) , B = [ α , k + α ) . Therefore, R 2 ( c, k ) ≥ 2 k + c + 4 α F or the upp er b ound, again we can lo ok at coloring of the interv al ( α , 2 k + c + 4 α ) . Sub case 3.1. α is colored red. ( α, α, c + 2 α ) c = ⇒ c + 2 α is blue . ( c + 2 α, c + 2 α , k + 2 c + 4 α ) k = ⇒ k + 2 c + 4 α is red . ( α, k + c + 3 α, k + 2 c + 4 α ) c = ⇒ k + c + 3 α is blue . 9 ( k + c 2 + 2 α, k + c 2 + 2 α, k + 2 c + 4 α ) c = ⇒ k + c 2 + 2 α is blue . Supp ose k − 2 c is red, ( α, k − 2 c, k − 2 c + α ) c = ⇒ k − c + α is blue . ( k − c + α , c + 2 α, 2 k + 3 α ) k = ⇒ 2 k + 3 α is red . ( α, 2 k + 3 α , 2 k + c + 4 α ) c = ⇒ 2 k + c + 4 α is blue . Th us ( k + c 2 + 2 α, k + c 2 + 2 α, 2 k + c + 4 α ) k is a blue solution, a contradiction. Supp ose k − 2 c is blue, ( c + 2 α, k − 2 c, 2 k − c + 2 α ) k = ⇒ 2 k − c + 2 α is red . ( k − c + α , k − c + α, 2 k − c + α ) c = ⇒ k − c + α is blue . ( k − c + α , c + α, 2 k + α ) k = ⇒ 2 k + 3 α is red . ( α, 2 k + 3 α , 2 k + c + 4 α ) c = ⇒ 2 k + c + 4 α is blue . Th us ( k + c 2 + 2 α, k + c 2 + 2 α, 2 k + c + 4 α ) k is a blue solution, a contradiction. Sub case 3.2. α is colored blue. ( α, α, k + 2 α ) k = ⇒ k + 2 α is red. ( k + 2 α, k + 2 α, 2 k + c + 4 α ) c = ⇒ 2 k + c + 4 α is blue. ( α, k + c + 3 α, 2 k + c + 4 α ) k = ⇒ k + c + 3 α is red. Supp ose k − 2 c is red, ( k − 2 c, k + 2 α, 2 k − c + 2 α ) c = ⇒ 2 k − c + 2 α is blue. ( k − c 2 + α, k − c 2 + α, 2 k − c + 2 α ) k = ⇒ k − c 2 + α is red. Th us ( k − c 2 + α, k − c 2 + α, k + 2 α ) c is a red solution, a contradiction. Supp ose k − 2 c is blue, ( k + c 2 + 2 α, k + c 2 + 2 α, 2 k + c + 4) k = ⇒ k + c 2 + 2 α is red. ( k − 2 c, 3 c + 4 α , 2 k + c + 4 α ) k = ⇒ 3 c + 4 α is red. ( c + 2 α, c + 2 α , 3 c + 4 α ) c = ⇒ c + 2 α is blue. ( c + 2 α, c + 2 α , k + 2 c + 4 α ) k = ⇒ k + 2 c + 4 α is red. Th us ( k + c 2 + 2 α, k + c 2 + 2 α, k + 2 c + 4 α ) c is a red solution, a contradiction. Thus, w e arriv e at a contradiction in b oth the sub cases. Therefore, R 2 ( c, k ) ≤ 2 k + c + 4 α . 10 4. Conclusion The theory of Rado num b ers in volv es numerous op en problems, whose complexity in- creases significantly with the n umber of colors. As a result, muc h of the literature has fo cused on the tw o-color case. In this setting, sev eral v arian ts—including disjunctiv e [5] and off-diagonal [4, 7, 9] and con tinuous [12, 13] Rado n umbers—hav e b een studied for v ari- ous class es of linear equations and systems. Nev ertheless, the existing work on off-diagonal Rado num bers is largely limited to homogeneous equations, with the non-homogeneous case remaining comparatively underexplored. In this work, we ev aluated the 2 -color off-diagonal Rado num b er, R 2 ( c, k ) , for x + y + c = z and x + y + k = z for b oth discrete and contin uous v ersions. Several directions remain op en for future inv estigation, including extending the present framework to broader families of equations encompassing b oth homogeneous and non-homogeneous cases, as w ell as generalizing the known results to colorings with more than tw o colors. References [1] Ra jat A dak, Y ash Bakshi, L Sunil Chandran, and Saraswati Girish Nanoti. 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