Odd-Power Impossibility and a Constructive Path Toward Fermat's Last Theorem

Odd-Power Impossibility and a Constructive Path Toward Fermat’s Last Theorem Walter W. Mayo December 7, 2025 Abstract This paper introduces a structural partition of the integers based on parity and factorization. Proper evens are defined as integers of the form 2k, and mixed evens as integers of the form 2k(2m+ 1), where k, m ∈Z. Odd integers of the form 2m+1 are further classified as class 1 odds when mis even, and class 3 odds when mis odd. This framework partitions the integers into four categories—proper evens, mixed evens, class 1 odds, and class 3 odds—and explores their properties in number-theoretic contexts. 1 Introduction Mathematicians study the properties of numbers, sets, and operators, as well as the relationships among them. These relationships often reveal deeper truths and contribute to the body of mathematical knowledge. Once a property is discovered, it becomes a foundation for further exploration. 2 Odd Integer Classification Consider the form 2m+ 1, where m∈Z. This expression generates all odd integers. •If m≡0 (mod 2), then 2m+ 1 ≡1 (mod 4): Class 1 odd. •If m≡1 (mod 2), then 2m+ 1 ≡3 (mod 4): Class 3 odd. Examples: •m= 2 ⇒2m+1=5⇒5≡1 (mod 4): class 1 odd. •m= 3 ⇒2m+1=7⇒7≡3 (mod 4): class 3 odd. 1 3 Even Integer Classification We define two types of even integers: Definition 1. Aproper even is an integer of the form 2k, where k∈Z+. Definition 2. Amixed even is an integer of the form 2k(2m+1), where k, m ∈ Z+. Examples: •23= 8: proper even. •22·3 = 12: mixed even. 4 Partition of the Integers This classification yields a four-part partition: 1. Proper evens: 2k 2. Mixed evens: 2k(2m+ 1) 3. Class 1 odds: 2m+ 1, m≡0 (mod 2) 4. Class 3 odds: 2m+ 1, m≡1 (mod 2) 5 p-adic Valuation Definition 3. The p-adic valuation of an integer r, denoted vp(r), is the exponent of the prime pin the prime factorization of r. Properties: •vp(mn)=vp(m)+vp(n) •vp(m±n) = min(vp(m), vp(n)) (assuming m±n= 0) Special Case: For p= 2: •If ris odd, then v2(r) = 0 •If ris even, then v2(r)≥1 6 Conclusion This framework provides a structured lens for analyzing integer properties, particularly in modular and parity-based investigations. The partitioning into proper evens, mixed evens, and two classes of odds supports deeper exploration of number-theoretic identities, valuations, and contradiction-based reasoning. 2 7 Some Observations Using Odd Forms 1. Let 2m+1 and 2r+1 be any two class 1 odds. Then (2m+1)+(2r+1) = 2(m+r+ 1). Since mand rare both assumed even, then the sum of any two class 1 odds is an odd multiple of 2. Then the 2 −adic valuation is 1. 2. Let 2m+ 1 and 2r+ 1 be any two class 3 odds. Then 2m+ 1 + 2r+ 1 = 2(m+r+ 1). Since mand rare both odd by definition of class 3 odds, the sum of any two class 3 odds has 2 −adic valuation 1. 3. Let 2m+ 1 be any class 1 odd and 2r+ 1 be any class 3 odd. Then 2m+ 1 + 2r+ 1 = 2(m+r+ 1). Since mis even and ris odd, it follows that: (a) If mis divisible by 2 but not 4 and ris a class 1 odd,then the 2 −adic valuation is >2. Pf:Since mis divisible by 2 but not 4,then m= 2hfor some odd integer h. Since ris a class 1 odd and 1 is a class 1 odd, then by (1) above, r+ 1 is divisible by 2 but not 4. So r+1 = 2jfor some odd integer j. It follows that 2(m+r+ 1) = 2(2h+ 2j) = 4(h+j) with hand jodd. This shows the 2 −adic valuation is >2. (b) If mis divisible by 2 but not 4 and ris a class 3 odd, then the 2−adic valuation is 2 exactly. Pf: Since mis divisible by 2 but not 4, then m= 2hwith hsome odd integer.And since ris a class 3 odd, then r+ 1 is divisible by 4. So r+ 1 = 4jwith jan integer. Then 2(m+r+ 1) = 2(2h+ 4j) = 4(h+ 2j), which shows the 2 −adic valuation is 2 exactly. (c) If mis divisible by 4 and ris a class 1 odd, then case (b) holds. Pf: Since mis divisible by 4, m= 4hfor some integer h. Since ris a class 1 odd, r+ 1 = 2jfor some odd integer j. Then 2(m+r+ 1) = 2(4h+ 2j) = 4(h+ 2j) which shows the 2 −adic valuation is 2 exactly. (d) If mis divisible by 4 and ris a class 3 odd, then the 2−adic valuation is ≥3. Pf: Since mis divisible by 4, m= 4hfor some integer h. Also, since ris a class 3 odd, r+ 1 is divisible by 4. So r= 2t+ 1 with tan odd integer since ris a class 3 odd. So r+ 1 = 2t+ 1 + 1 = 2(t+ 1) = 4jfor some integer j. Then 2(m+r+ 1) = 2(4h+ 4j) = 8(h+j) with 2 −adic valuation ≥3. 4. Let 2m+1 and 2r+1 be any two distinct class 1 odds (class 3 odds). Then 2m+1−(2r+1) = 2(m−r) has 2−adic valuation ≥2. For m, and rboth even with distinct 2 −adic valuations , the 2 −adic valuation of 2(m−r) will be v2(m) + 1 or v2(r) + 1 whichever is less. For v2(m) = v2(r), then v2(m−r) is ≥v2(m) +2. 3 Pf: Let 2m+ 1 and 2r+ 1 be distinct class 1 odds, then mand rare both even by definition. Let m= 2aA, with a > 0 and Aodd. Let r= 2bB, with b>0 and Bodd. Without loss of generality, let a > b. Then 2(m−r) = 2(2aA−2bB)=2b+1(2a−bA−B) and b+ 1 is v2(r) + 1. For mand rdistinct class 1 odds (class 3 odds), the 2 −adic valuation of 2(m−r) is ≥3. If one of mand ris class 1 and the other is class 3, then the 2 −adic valuation of 2(m−r) is 2 exactly. Pf: If mand rare both class 1 odds, then m−ris divisible by 4 so 2(m−r) is divisible by 8, so 2−adic valuation is ≥3. The same line of reasoning holds for mand r as class 3 odds. Now, if mis a class 1 odd and ris class 3, then m−ris divisible by 2 but not 4. So in this case v2[2(m−r)] is 2 exactly. 5. If 2m+ 1 is a class 1 odd, then (2m+ 1)nis a class 1 odd, nany positive integer. Pf:(2m+ 1)n= (2m)n+n(2m)n−1+... +n(2m) + 1. Since mis even, we may factor 2 from each term except the last to get the form 2r+ 1 where ris even. This is the form of a class 1 odd. 6. If 2m+ 1 is a class 3 odd, then (2m+ 1)nis a class 1 odd if nis even and a class 3 odd if ris odd. Pf: If nis even, then (2m+ 1)n= (2m)n+n(2m)n−1+... +n(2m) + 1, and since nis even, a 2 factors out of each term except the last. Each term except the last is still even since each term contains 2m. So, the form is 2r+ 1 with reven, which is the form of a class 1 odd. If nis odd, then (2m+ 1)n= (2m)n+n(2m)n−1+... +n(2m) + 1. Factor out a 2 to get 2(2n−1mn+... +nm) + 1 which is the form 2(2M+nm) + 1, a class 3 odd. It follows that the product of any number of class 1 odds is a class 1 odd. The product of an even number of class 3 odds is a class 1 odd, and the product of an odd number of class 3 factors will be a class 3 odd. The product of a class 1 odd and a class 3 odd will be a class 3 odd since (2m+ 1)(2r+ 1) = 2m2r+ 2m+ 2r+ 1 = 2(m2r+m+r) + 1. The term m2r+m+ris odd, so this is the form 2M+ 1 with Modd. So this product is a class 3 odd. 4 8 Observations of Proper and mixed Evens 1. Let 2kand 2jbe distinct proper evens. Then their sum will always be a mixed even. Pf: Since the evens are distinct, we may let k > j without loss of generality. Then 2k+ 2j= 2j(2k−j+ 1), which is the form of a mixed even. If k=j, the evens are not distinct, and the sum will be 2k+ 2k= 2k+1. Remark: If k−j= 1, then the odd factor will always be 3. A number of this form is the center of intervals [ 2k,2k+1 ]. The center will be 2k−1(3). I can arrange a difference so that I obtain a positive answer. 2. Let 2kand 2jbe distinct proper evens with k > j. Then 2k−2j= 2j(2k−j−1) is a proper even if k−j= 1 and a mixed even if k−j > 1. 3. Let 2k(2m+ 1) and 2j(2r+ 1) be distinct mixed evens, then their sum is mixed even if k > j or k < j. If k=j, then consider cases. Pf: Let k > j. Then write the sum as 2j[2k−j(2m+ 1) + (2r+ 1)] which is the form of a mixed even. If j > k, then write the sum as 2k[(2m+1)+2j−k(2r+1)] which is the form of a mixed even. Let k=j, and consider the following cases for the sum 2k[2m+1+2r+1] = 2k+1[m+r+1] : Case (a), with mand rboth even: In the expression 2k+1[m+r+ 1], if mand rare both even, then [m+r+ 1] will be an odd factor, so the expression is the form of a mixed even. Case (b), with mand r both odd: It is clear the factor [m+r+ 1] will be odd, so again, it is a mixed even. Case (c), with mand ropposite parity. In this case, the expression [m+r+ 1] is even. We need additional information or additional assumptions to determine whether we have a proper or mixed even. 4. Consider the difference of distinct mixed evens, 2k(2m+1)−2j(2r+ 1). With k > j and (2m+1), (2r+1) distinct, then 2j[2k−j(2m+1)−(2r+1)] is a mixed even. With k=j, and (2m+ 1) and (2r+ 1) distinct, then the result is 2k+1[m−r]. If mand rare opposite parity, then this is a mixed even. If mand rare the same parity, one needs more information or additional assumptions. 5. Consider the difference of a proper and mixed even, with 2k>2j(2r+ 1). The difference, 2j[2k−j−(2r+ 1)], is a mixed even. 5 6. For the difference 2k(2m+1)−2jconsider (a) k=j,(b) k > j,(c) k < j. (a) If k=j, 2k[2m+1−1] = 2k+1[m], and mis odd, then mixed even. If mis even, then one needs more information. But, one can say the 2−adic valuation is > k + 1. (b) If k > j, then 2j[2k−j(2m+ 1) −1] is a mixed even. (c) If k < j, then 2k[(2m+ 1) −2j−k] is a mixed even. 7. Now consider the sum of a proper and mixed even, 2k+ 2j(2m+ 1). (a) If k=j, then write 2k[1 + 2m+ 1] = 2k+1[m+ 1]. If mis even, then the sum is a mixed even. If m= 1, then the sum is a proper even. If mis a class 1 odd >1, then the sum is mixed even since the sum of class 1 odds is an odd multiple of 2. For ma class 3 odd of the form 2d−1, with d≥2, then it is a proper even. For ma class 3 odd not expressible as 2d−1, then the result is a mixed even. 9 A Pythagorean Criterion Consider two odd integers, yand z. It is known that for the solutions to the Pythagorean equation x2+y2=z2(1) there exist integers aand bsuch that z=a2+b2and y=a2−b2. And since z and yare both odd, then aand bare opposite parity. If ais even, then zmust be a class 1 odd and ymust be a class 3 odd. If bis even, then zand ymust both be class 1 odds. Pf: If ais even, then z−b2=a2. Now b2is a class 1 odd and z−b2is an even divisible 4, so zis a class 1 odd. y+b2=a2is an even divisible by 4, so ymust be a class 3 odd. If bis even, z−a2is divisible by 4, and a2−yis divisible by 4, so zand yare both class 1 odds. 10 The Even Part of an Odd Integer In the form 2m+ 1, it might be helpful to think of 2mas the even part of the form. When an odd integer is raised to a power r, the 2 −adic valuation of the even part, 2m, does not change if the power is odd. But, if ris an even power, the 2 −adic valuation of 2mwill increase. 6 11 The Pythagorean Theorem The Pythagorean Theorem dates back to ancient times, around 600 B.C. It is the statement that for any right triangle with side lengths aand band hypotenuse length c, the relationship a2+b2=c2(2) always holds. We are seeking to find a set of triples a, b, c such that the substitution of these numbers into the equation gives a true statement. A well-known solution is 3,4,5. This gives 32+ 42= 52. For this set, the numbers are relatively prime in pairs. That means no two of them have any prime number in common. Pairwise relatively prime solutions are called primitive solutions. Other examples of primitive solutions are (5,12,13), (8,15,17), and (7,24,25). Once a primitive solution is found, an infinite number of non-primitive solutions can be obtained. For example, in the set 3,4,5, one can multiply each number by n>1 to get (3n)2+ (4n)2= (5n)2, which for each choice of n > 1, will obtain a nonprimitive solution. To find primitive solutions, let aand bbe positive integers with a>b, gcd(a, b) = 1, aand bwith opposite parity. Then the triple (x, y, z) given by x= 2ab, y =a2−b2, z =a2+b2(3) is a primitive solution. This establishes a one-to-one correspondence between the pairs (a, b) to satisfy the condition and the set of primitive solutions. As an exercise, investigate the Pythagorean equation using the forms 2k,2j(2m+ 1), and(2r+ 1). The possibilities for a solution set must look like one of the following: x= 2k, y = (2m+ 1), z = (2r+ 1) (4) x= 2k(2l+ 1), y = (2m+ 1), z = (2r+ 1) (5) The first set of equations has a proper even; the second set has an improper even. Lemma: The sum and difference of two odd integers cannot both be divisible by 4. So, if the sum is divisible by 4, then the difference will be divisible by 2 but not 4. If the difference is divisible by 4, then the sum will be divisible by 2 but not 4. Pf : The sum and difference of two odd integers is even. Let aand bbe two odd integers. Suppose that a+band a−bare divisible by 4, then their sum is divisible by 4. So, (a+b)+(a−b) = 2a= 4hfor some integer h. But ais odd, so 2ais an odd multiple of 2. That is a contradiction. That shows that either the sum or the difference is divisible by 4, not both. In the Pythagorean equation, x2+y2=z2,xand yare not both odds. If xand ywere both odd, then the sum of the two class 1 odds (any odd square is a class 1 odd) is equal to an even number squared. But an even square has 2 −adic valuation greater than or equal to 2, but the sum of two class 1 odds has 2 −adic valuation equal to 1. We conclude that xand yhave opposite parity. Without loss of generality, 7 let xbe even and write: (2k)2= (2r+ 1)2−(2m+ 1)2(6) (2k)2= 4(r−m)(r+m+ 1) (7) It follows from equation (7) that rand mmust have opposite parity. If both are even or both are odd, then one of the factors (r−m), (r+m+ 1) will be odd, which is impossible since the LHS of the equation is a proper even. This means that (r−m) = 1 and so r=m+1). It also holds here that rmust be even. If r were odd, then r−1 = mand the factor (r+m+1) = r+(r−1)+1 = 2r) would be an odd multiple of 2 and 8 on RHS is not a square. One can also suppose that ris a proper even. This establishes that in the Pythagorean equation, as expressed here, ymust be a class 3 odd, and zmust be a class 1 odd. A particular case is x= 4, y= 3, z= 5. Using these results, write the equation(7) as (2k)2= 23(m+ 1) (8) So we know here that k= 1 is impossible; if k= 2, then m= 1 and r= 2. This gives the set (3,4,5). If k= 3, then m= 7 and so r= 8. This gives the set (8,15,17). For every k > 2, we can find an mthat works for xa proper even greater than 2. Now consider the case with xa mixed even and write [2k(2l+ 1)]2= (2r+ 1)2−(2m+ 1)2= 4(r−m)(r+m+ 1) (9) It is clear that k≥2. Since (2r+ 1) and (2m+ 1) are relatively prime. The factors (r−m) and (m+r+ 1) are relatively prime. Pf : Suppose there is a prime pthat divides both (r−m) and (r+m+ 1). Then pwill divide their sum and difference. (r−m)=hp, (r+m+ 1) = ip (10) Now add the two equations in (10) to get (r−m) + (r+m+ 1) = (h+i)p(11) If one subtracts the two equations in (10), it is clear that palso divides (2m+1) The sum on the LHS is 2r+ 1, so pdivides 2r+ 1. Now subtract the two equations in (10) to get (r+m+ 1) −(r−m)=2m+ 1 = (i−h)p(12) The sum on the LHS is 2m+1, so pdivides 2m+1. But 2r+1 and 2m+1 were supposed to be relatively prime. We conclude that the factors r−mand r+m+1 are relatively prime. This means they are squares. So write (r−m) = b2and (r+m+1) = a2. If one adds and subtracts the two equations, (2r+1) = a2+b2 8 and (2m+ 1) = a2−b2, where aand bhave opposite parity since 2r+ 1 and 2m+ 1 are both odd. Now, write these equations as 2r=a2+b2−1 (13) 2m=a2−b2−1 (14) From the last equation, with aeven and bodd, rmust be even and mmust be odd. So 2r+ 1 must be a class 1 odd, and 2m+ 1 is a class 3 odd. If a is odd and bis even, then rand mmust both be even. So both 2r+ 1 and 2m+ 1 are class 1 odds. It is clear that it is impossible to have both odds in the Pythagorean equation as class 3 odds. We also see that when the odds are class 1 and class 3, the larger odd must be a class 1 odd. This follows from the fact that a2+b2−1 is always greater than a2−b2−1 if aand bare non-zero. 12 Pierre Fermat’s Question Once one is familiar with the Pythagorean equation x2+y2=z2, it is natural to question whether there exist integers that might satisfy an equation like x3+y3=z3. It seems such a minor modification of the Pythagorean equation (just changing the exponent from 2 to 3), and with an infinite sea of integers to consider, one feels almost certain that there is at least one solution set. From the Pythagorean Theorem, we learn that every right triangle with sides a,b, and hypotenuse csatisfies equation a2+b2=z2. From elementary mathematics, we learn that the area of any right triangle is given by A= (1/2)ab, where aand bare the base and height of the triangle. The French mathematician Pierre Fermat (1607-1665) wondered if the area of any triangle was a square number. He answered this question in the negative. He then stated: “The area of a rational triangle cannot be a square number. The proof of this theorem I have reached with elaborate and ardent study.” [1]. This would not be the last time Fermat would make such a claim. He would repeat this with respect to one of the most famous conjectures in the history of mathematics: The Last Theorem of Fermat. Along the margin of his copy of the works of Diaphantus, Fermat wrote“It is impossible to write a cube as the sum of two cubes, a fourth power as the sum of two fourth powers, and in general any power beyond the second as the sum of two similar powers. For this, I have discovered a truly wonderful proof, but the margin is too small to contain it.” [2]. This is the famous Fermat’s Last Theorem. This means there is no positive integer solution to the equation xn+yn=zn, n > 2 (15) For the case n= 4, Fermat had a proof. This follows from his work that the area of a right triangle cannot be a square. It is also likely that he had a proof for the case n= 3. It is known that he gave this case as a challenge to other mathematicians; he probably would not have done so unless he himself knew a solution. The first known published proof for the case n= 3 was written by Leonard Euler. A proof of the general theorem by Fermat was never found. 9 = (2r+ 1)p+ (2m+ 1)p= 2pk(2L+ 1)p(52) Now divide through by (2L+ 1)pto get (2r+ 1) + (2m+ 1) + [2aA(2r+1)−2bB(2m+ 1)]/(2L+ 1)p= 2pk (53) On the LHS of equation (53), the sum (2r+ 1) + 2m+ 1) is equal to the RHS of equation (53). Subtract 2pk from both sides of equation (53) to get [2aA(2r+1)−2bB(2m+ 1)]/(2L+ 1)p= 0 (54) and 2aA(2r+1)−2bB(2m+ 1) = 0 (55) It is clear here that 2aand 2bmust be equal. After canceling some terms, rearrange equation (55) to get A(2r+ 1) = B(2m+ 1) (56) Since (2r+ 1) and (2m+ 1) are relatively prime to each other and relatively prime to (2L+ 1), and Aand Bare integers, this equality might be untenable. To have equality in equation (56), one could have A= (2m+1) and B= (2r+1). But then there is a contradiction using equations (47) and (48). We substitute, respectively, (2m+ 1) and (2r+ 1) for Aand Bin equations (47) and (48) and subtract equation (48) from (47) to get 2a(2m+ 1) + 2a(2r+ 1) = (2r+ 1)p−1−(2m+ 1)p−1(57) 2a[(2r+ 1) + (2m+ 1)] = (2r+ 1)p−1−(2m+ 1)p−1(58) . But (2r+ 1) + (2m+ 1) = 2pk on LHS of (58). So, the LHS is a pure even; the RHS will have an odd factor. That means that the Fermat equation (38) fails using the sum of odds under the given conditions. The same idea works even if the sum of 2r+ 1 and 2m+1 has an odd factor in the RHS with the even 2pk. In that case, start with 2r+ 1 + 2m+ 1 = 2pkD(59) where Dis an odd consisting of one or more odd primes of 2L+ 1. That work will be shown after considering the difference of odds equal to an even. Difference of Odds Now consider the Fermat equation with the difference of two odds. (2r+ 1)p−(2m+ 1)p= 2pk(2L+ 1)p(60) Assume that the difference (2r+ 1) −(2m+ 1) gives 2pk exactly. 16 16 Difference of two odds equal to a pure power of two Assume a primitive solution to (2r+ 1)p−(2m+ 1)p= 2pk(2L+ 1)p, with (2r+1)−(2m+ 1) = 2pk and podd prime,gcd(2r+ 1,2m+ 1) = 1. For odd a, b and odd prime p, v2(ap−bp)=v2(a−b) + v2(a+b). Setting a= 2r+ 1, b= 2m+ 1, we obtain v2(2r+ 1)p−(2m+ 1)p=v2(2r+1)−(2m+ 1)+v2(2r+ 1) + (2m+ 1). By assumption, v2(2r+1)−(2m+1)=pk, and since (2r+1)+(2m+1) = 2(r+m+ 1), we have v22(r+m+ 1)≥1. Hence v2(2r+ 1)p−(2m+ 1)p≥pk + 1. On the other hand, v22pk(2L+ 1)p=pk. This contradiction shows that no primitive solution exists when the difference of the two odds is a pure power of two. 17 Valuation Lemma and Corollaries We now unify the previous cases (sum of odds, difference of odds, with or without odd factors) under a single valuation principle, showing that each form of the Fermat equation fails by the same contradiction. Proof. This is the standard lifting-the-exponent (LTE) identity for the prime q= 2 (the 2-adic valuation) with odd exponent p, applied to odd integers a, b. Under these conditions, LTE yields v2(ap−bp)=v2(a−b) + v2(a+b). 17 Corollary 1: Sum of odds Suppose (2r+ 1)p+ (2m+ 1)p= 2pk(2L+ 1)p, with (2r+ 1) + (2m+ 1) = 2pk or 2pkVfor some odd V. Note that (2r+ 1)p+ (2m+ 1)p= (2r+ 1)p−[−(2m+ 1)]p, and since pis odd, −(2m+ 1) is odd, so the lemma applies. Then v2(2r+ 1)p+ (2m+ 1)p≥pk + 1, while the right-hand side has valuation exactly pk. Contradiction. Corollary 2: Difference of odds (pure power of two) Suppose (2r+ 1)p−(2m+ 1)p= 2pk(2L+ 1)p, with (2r+ 1) −(2m+ 1) = 2pk. Then v2(2r+1)p−(2m+1)p=pk+v2((2r+1)+(2m+1)) = pk+1+v2(r+m+1) ≥pk+1, contradicting the valuation of the right-hand side. Corollary 3: Difference of odds with odd factor Suppose (2r+ 1)p−(2m+ 1)p= 2pk(2L+ 1)p, with (2r+ 1) −(2m+ 1) = 2pkV,Vodd. Then v2(2r+ 1)p−(2m+ 1)p=pk +v2((2r+ 1) + (2m+ 1)) ≥pk + 1, while the right-hand side has valuation pk. Contradiction again. Conclusion In all cases, the left-hand side acquires at least one more factor of 2 than the right-hand side. This valuation mismatch rules out primitive solutions for these forms of the Fermat equation. Together with earlier parity classifications, this shows how 2-adic arguments systematically eliminate the possibility of solutions when nis an odd prime, aligning with Fermat’s Last Theorem. 18 Final Remarks Through a systematic classification of integers into proper evens, mixed evens, and odd classes, combined with 2-adic valuation analysis, we have shown that every attempted form of the Fermat equation fails under parity and valuation constraints. Whether expressed as a sum of odds, a difference of odds, or with additional odd factors, the left-hand side always acquires at least one extra factor of 2 beyond the right-hand side. This persistent mismatch demonstrates the impossibility of primitive solutions for odd prime exponents. These observations suggest that parity and valuation arguments provide a natural sieve against potential solutions, and they illustrate how elementary number-theoretic tools can be organized into a modular framework that aligns with Fermat’s Last Theorem. While the modern proof relies on deep connections to elliptic curves and modular forms, the approach here highlights how structural properties of integers themselves can be leveraged to eliminate candidate solutions in a transparent and teachable way. 19