{
  "index": "1976-A-3",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "A-3. Find all integral solutions of the equation\n\\[\n\\left|p^{\\prime}-q^{s}\\right|=1,\n\\]\nwhere \\( p \\) and \\( q \\) are prime numbers and \\( r \\) and \\( s \\) are positive integers larger than unity. Prove that there are no other solutions.",
  "solution": "A-3.\nWe show that the only solutions are given by \\( 3^{2}-2^{3}=1 \\), i.e.. \\( (p, r, q, s)=(3,2,2,3) \\) or \\( (2,3,3,2) \\).\nClearly ether \\( p \\) or \\( q \\) is 2 . Suppose \\( q=2 \\). Then \\( p \\) is an odd prime with \\( p^{\\prime} \\pm 1=2^{3} \\). If \\( r \\) is odd, \\( \\left(p^{\\prime} \\pm 1\\right) /(p \\pm 1) \\) is the odd integer \\( p^{\\prime-1} \\mp p^{\\prime-2}+p^{\\prime-3} \\mp p^{\\prime-4}+\\cdots+1 \\). which is greater than 1 since \\( r>1 \\); this contradicts the fact that \\( 2^{\\prime \\prime} \\) has no such factor.\n\nNow we try \\( r \\) as an even integer \\( 2 t \\). Then \\( p^{\\prime}+1=2^{3} \\) leads to\n\\[\n2^{\\prime}=\\left(p^{\\prime}\\right)^{2}+1=(2 n+1)^{2}+1=4 n^{2}+4 n+2\n\\]\nwhich is impossible since \\( 42^{5} \\) for \\( s>1 \\) and \\( 4 \\times\\left(4 n^{2}+4 n+2\\right) \\).\nAlso \\( r=2 t \\) and \\( p^{\\prime}-1=2^{\\prime} \\) leads to \\( \\left(p^{\\prime}\\right)^{2}-1=(2 n+1)^{2}-1=4 n^{2}+4 n=4 n(n+1)=2^{3} \\). Since either \\( n \\) or \\( n+1 \\) is odd. this is only possible for \\( n=1, s=3, p=3 \\), and \\( r=2 \\).",
  "vars": [
    "p",
    "q",
    "r",
    "s",
    "t",
    "n"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "p": "primebase",
        "q": "secondprime",
        "r": "exponentone",
        "s": "exponenttwo",
        "t": "halflength",
        "n": "auxinteger"
      },
      "question": "A-3. Find all integral solutions of the equation\n\\[\n\\left|primebase^{\\prime}-secondprime^{exponenttwo}\\right|=1,\n\\]\nwhere \\( primebase \\) and \\( secondprime \\) are prime numbers and \\( exponentone \\) and \\( exponenttwo \\) are positive integers larger than unity. Prove that there are no other solutions.",
      "solution": "A-3.\nWe show that the only solutions are given by \\( 3^{2}-2^{3}=1 \\), i.e., \\( (primebase, exponentone, secondprime, exponenttwo)=(3,2,2,3) \\) or \\( (2,3,3,2) \\).\nClearly either \\( primebase \\) or \\( secondprime \\) is 2. Suppose \\( secondprime = 2 \\). Then \\( primebase \\) is an odd prime with \\( primebase^{\\prime} \\pm 1 = 2^{3} \\). If \\( exponentone \\) is odd, \\( \\left( primebase^{\\prime} \\pm 1 \\right)/(primebase \\pm 1) \\) is the odd integer \\( primebase^{\\prime-1} \\mp primebase^{\\prime-2} + primebase^{\\prime-3} \\mp primebase^{\\prime-4} + \\cdots + 1 \\), which is greater than 1 since \\( exponentone > 1 \\); this contradicts the fact that \\( 2^{\\prime \\prime} \\) has no such factor.\n\nNow we try \\( exponentone \\) as an even integer \\( 2\\,halflength \\). Then \\( primebase^{\\prime} + 1 = 2^{3} \\) leads to\n\\[\n2^{\\prime} = \\left( primebase^{\\prime} \\right)^{2} + 1 = (2\\,auxinteger + 1)^{2} + 1 = 4\\,auxinteger^{2} + 4\\,auxinteger + 2\n\\]\nwhich is impossible since \\( 42^{5} \\) for \\( exponenttwo > 1 \\) and \\( 4 \\times \\left( 4\\,auxinteger^{2} + 4\\,auxinteger + 2 \\right) \\).\nAlso \\( exponentone = 2\\,halflength \\) and \\( primebase^{\\prime} - 1 = 2^{\\prime} \\) leads to \\( \\left( primebase^{\\prime} \\right)^{2} - 1 = (2\\,auxinteger + 1)^{2} - 1 = 4\\,auxinteger^{2} + 4\\,auxinteger = 4\\,auxinteger(auxinteger + 1) = 2^{3} \\). Since either \\( auxinteger \\) or \\( auxinteger + 1 \\) is odd, this is only possible for \\( auxinteger = 1, exponenttwo = 3, primebase = 3 \\), and \\( exponentone = 2 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "p": "longitude",
        "q": "pineapple",
        "r": "carousel",
        "s": "strawhat",
        "t": "buttercup",
        "n": "umbrellax"
      },
      "question": "A-3. Find all integral solutions of the equation\n\\[\n\\left|longitude^{\\prime}-pineapple^{strawhat}\\right|=1,\n\\]\nwhere \\( longitude \\) and \\( pineapple \\) are prime numbers and \\( carousel \\) and \\( strawhat \\) are positive integers larger than unity. Prove that there are no other solutions.",
      "solution": "A-3.\nWe show that the only solutions are given by \\( 3^{2}-2^{3}=1 \\), i.e.. \\( (longitude, carousel, pineapple, strawhat)=(3,2,2,3) \\) or \\( (2,3,3,2) \\).\nClearly ether \\( longitude \\) or \\( pineapple \\) is 2 . Suppose \\( pineapple=2 \\). Then \\( longitude \\) is an odd prime with \\( longitude^{\\prime} \\pm 1=2^{3} \\). If \\( carousel \\) is odd, \\( \\left(longitude^{\\prime} \\pm 1\\right) /(longitude \\pm 1) \\) is the odd integer \\( longitude^{\\prime-1} \\mp longitude^{\\prime-2}+longitude^{\\prime-3} \\mp longitude^{\\prime-4}+\\cdots+1 \\), which is greater than 1 since \\( carousel>1 \\); this contradicts the fact that \\( 2^{\\prime \\prime} \\) has no such factor.\n\nNow we try \\( carousel \\) as an even integer \\( 2 buttercup \\). Then \\( longitude^{\\prime}+1=2^{3} \\) leads to\n\\[\n2^{\\prime}=\\left(longitude^{\\prime}\\right)^{2}+1=(2 umbrellax+1)^{2}+1=4 umbrellax^{2}+4 umbrellax+2\n\\]\nwhich is impossible since \\( 42^{5} \\) for \\( strawhat>1 \\) and \\( 4 \\times\\left(4 umbrellax^{2}+4 umbrellax+2\\right) \\).\nAlso \\( carousel=2 buttercup \\) and \\( longitude^{\\prime}-1=2^{\\prime} \\) leads to \\( \\left(longitude^{\\prime}\\right)^{2}-1=(2 umbrellax+1)^{2}-1=4 umbrellax^{2}+4 umbrellax=4 umbrellax(umbrellax+1)=2^{3} \\). Since either \\( umbrellax \\) or \\( umbrellax+1 \\) is odd, this is only possible for \\( umbrellax=1, strawhat=3, longitude=3 \\), and \\( carousel=2 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "p": "compositeone",
        "q": "compositetwo",
        "r": "logarithm",
        "s": "antilogar",
        "t": "singular",
        "n": "negative"
      },
      "question": "A-3. Find all integral solutions of the equation\n\\[\n\\left|compositeone^{\\prime}-compositetwo^{antilogar}\\right|=1,\n\\]\nwhere \\( compositeone \\) and \\( compositetwo \\) are prime numbers and \\( logarithm \\) and \\( antilogar \\) are positive integers larger than unity. Prove that there are no other solutions.",
      "solution": "A-3.\nWe show that the only solutions are given by \\( 3^{2}-2^{3}=1 \\), i.e.. \\( (compositeone, logarithm, compositetwo, antilogar)=(3,2,2,3) \\) or \\( (2,3,3,2) \\).\nClearly ether \\( compositeone \\) or \\( compositetwo \\) is 2 . Suppose \\( compositetwo=2 \\). Then \\( compositeone \\) is an odd prime with \\( compositeone^{\\prime} \\pm 1=2^{3} \\). If \\( logarithm \\) is odd, \\( \\left(compositeone^{\\prime} \\pm 1\\right) /(compositeone \\pm 1) \\) is the odd integer \\( compositeone^{\\prime-1} \\mp compositeone^{\\prime-2}+compositeone^{\\prime-3} \\mp compositeone^{\\prime-4}+\\cdots+1 \\), which is greater than 1 since \\( logarithm>1 \\); this contradicts the fact that \\( 2^{\\prime \\prime} \\) has no such factor.\n\nNow we try \\( logarithm \\) as an even integer \\( 2 singular \\). Then \\( compositeone^{\\prime}+1=2^{3} \\) leads to\n\\[\n2^{\\prime}=\\left(compositeone^{\\prime}\\right)^{2}+1=(2 negative+1)^{2}+1=4 negative^{2}+4 negative+2\n\\]\nwhich is impossible since \\( 42^{5} \\) for \\( antilogar>1 \\) and \\( 4 \\times\\left(4 negative^{2}+4 negative+2\\right) \\).\nAlso \\( logarithm=2 singular \\) and \\( compositeone^{\\prime}-1=2^{\\prime} \\) leads to \\( \\left(compositeone^{\\prime}\\right)^{2}-1=(2 negative+1)^{2}-1=4 negative^{2}+4 negative=4 negative(negative+1)=2^{3} \\). Since either \\( negative \\) or \\( negative+1 \\) is odd, this is only possible for \\( negative=1, antilogar=3, compositeone=3 \\), and \\( logarithm=2 \\)."
    },
    "garbled_string": {
      "map": {
        "p": "qzxwvtnp",
        "q": "hjgrksla",
        "r": "plmnbvcx",
        "s": "qwertyui",
        "t": "asdfghjk",
        "n": "zxcvbnml"
      },
      "question": "A-3. Find all integral solutions of the equation\n\\[\n\\left|qzxwvtnp^{\\prime}-hjgrksla^{qwertyui}\\right|=1,\n\\],\nwhere \\( qzxwvtnp \\) and \\( hjgrksla \\) are prime numbers and \\( plmnbvcx \\) and \\( qwertyui \\) are positive integers larger than unity. Prove that there are no other solutions.",
      "solution": "A-3.\nWe show that the only solutions are given by \\( 3^{2}-2^{3}=1 \\), i.e.. \\( (qzxwvtnp, plmnbvcx, hjgrksla, qwertyui)=(3,2,2,3) \\) or \\( (2,3,3,2) \\).\nClearly ether \\( qzxwvtnp \\) or \\( hjgrksla \\) is 2 . Suppose \\( hjgrksla=2 \\). Then \\( qzxwvtnp \\) is an odd prime with \\( qzxwvtnp^{\\prime} \\pm 1=2^{3} \\). If \\( plmnbvcx \\) is odd, \\( \\left(qzxwvtnp^{\\prime} \\pm 1\\right) /(qzxwvtnp \\pm 1) \\) is the odd integer \\( qzxwvtnp^{\\prime-1} \\mp qzxwvtnp^{\\prime-2}+qzxwvtnp^{\\prime-3} \\mp qzxwvtnp^{\\prime-4}+\\cdots+1 \\), which is greater than 1 since \\( plmnbvcx>1 \\); this contradicts the fact that \\( 2^{\\prime \\prime} \\) has no such factor.\n\nNow we try \\( plmnbvcx \\) as an even integer \\( 2 asdfghjk \\). Then \\( qzxwvtnp^{\\prime}+1=2^{3} \\) leads to\n\\[\n2^{\\prime}=\\left(qzxwvtnp^{\\prime}\\right)^{2}+1=(2 zxcvbnml+1)^{2}+1=4 zxcvbnml^{2}+4 zxcvbnml+2\n\\]\nwhich is impossible since \\( 42^{5} \\) for \\( qwertyui>1 \\) and \\( 4 \\times\\left(4 zxcvbnml^{2}+4 zxcvbnml+2\\right) \\).\nAlso \\( plmnbvcx=2 asdfghjk \\) and \\( qzxwvtnp^{\\prime}-1=2^{\\prime} \\) leads to \\( \\left(qzxwvtnp^{\\prime}\\right)^{2}-1=(2 zxcvbnml+1)^{2}-1=4 zxcvbnml^{2}+4 zxcvbnml=4 zxcvbnml(zxcvbnml+1)=2^{3} \\). Since either \\( zxcvbnml \\) or \\( zxcvbnml+1 \\) is odd, this is only possible for \\( zxcvbnml=1, qwertyui=3, qzxwvtnp=3 \\), and \\( plmnbvcx=2 \\)."
    },
    "kernel_variant": {
      "question": "Let $k\\ge 3$ and let $p_{1},\\dots ,p_{k}$ be pairwise distinct odd primes.  \nFor every $i$ fix an exponent $a_{i}\\ge 2$ and set  \n\\[\nN_{i}=p_{i}^{\\,a_{i}}\\qquad(1\\le i\\le k).\n\\]  \nAssume that for every pair of indices $1\\le i<j\\le k$ we have  \n\\[\n\\lvert N_{i}-N_{j}\\rvert \\;=\\; 2^{\\,b_{ij}},\\qquad b_{ij}\\ge 2. \n\\tag{$\\star$}\n\\]  \nProve that no such family $\\bigl(p_{i},a_{i}\\bigr)_{i=1}^{k}$ exists; that is, it is impossible to find three (or more) distinct odd prime powers of exponent at least $2$ whose pairwise differences are powers of two with exponent at least $2$.\n\n(The condition $b_{ij}\\ge 2$ rules out the powers of $2$ themselves; every $N_{i}$ is odd.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Assume, for a contradiction, that $k\\ge 3$ and that the odd prime powers  \n\\[\nN_{1}<N_{2}<\\dots <N_{k},\\qquad N_{i}=p_{i}^{\\,a_{i}}\\;(a_{i}\\ge 2),\n\\]\nsatisfy $(\\star)$.  Put\n\\[\nd_{ij}=N_{j}-N_{i}=2^{\\,b_{ij}}\\qquad\n\\bigl(b_{ij}\\ge 2,\\ 1\\le i<j\\le k\\bigr).\n\\tag{1}\n\\]\n\n--------------------------------------------------------------------\nStep 1.  (Consecutive gaps coincide.)  \n\nTake any triple of indices $i<j<\\ell$ and write\n\\[\nd_{ij}=2^{\\,x},\\quad d_{j\\ell}=2^{\\,y},\\quad d_{i\\ell}=2^{\\,z},\n\\qquad x,y,z\\ge 2.\n\\tag{2}\n\\]\nThen\n\\[\n2^{\\,x}+2^{\\,y}=2^{\\,z}.\n\\tag{3}\n\\]\nAssume without loss of generality $x\\le y$.  Dividing by\n$2^{\\,x}$ yields\n\\[\n1+2^{\\,y-x}=2^{\\,z-x}.\n\\tag{4}\n\\]\nIf $y-x\\ge 1$ the left–hand side of (4) is {\\em odd} whereas\nthe right–hand side is an {\\em even} power of two – impossible.\nHence $y-x=0$, so $x=y$ and (3) gives $z=x+1$.  Consequently\n\\[\nd_{ij}=d_{j\\ell}\\qquad(\\forall\\,i<j<\\ell)\n\\tag{5}\n\\]\nand\n\\[\nd_{i\\ell}=2\\,d_{ij}.\n\\tag{6}\n\\]\n\n--------------------------------------------------------------------\nStep 2.  (The numbers form an arithmetic progression.)  \n\nPut $t:=b_{12}$.  By (5) every consecutive gap equals $2^{t}$:\n\\[\nd_{r,r+1}=2^{\\,t}\\qquad(1\\le r\\le k-1).\n\\tag{7}\n\\]\nThus\n\\[\nN_{r}=N_{1}+(r-1)\\,2^{\\,t}\\qquad(1\\le r\\le k).\n\\tag{8}\n\\]\nIn particular,\n\\[\nN_{2}-N_{1}=2^{\\,t},\\qquad\nN_{3}-N_{2}=2^{\\,t},\\qquad\nN_{3}-N_{1}=2^{\\,t+1}.\n\\tag{9}\n\\]\nWrite\n\\[\nN_{1}=p^{a},\\;N_{2}=q^{b},\\;N_{3}=r^{c},\n\\quad p,q,r\\ \\text{odd primes},\\;\na,b,c\\ge 2.\n\\tag{10}\n\\]\nThen\n\\[\nq^{b}-p^{a}=2^{\\,t},\\qquad\nr^{c}-q^{b}=2^{\\,t},\\qquad\nr^{c}-p^{a}=2^{\\,t+1}.\n\\tag{11}\n\\]\n\n--------------------------------------------------------------------\nStep 3.  (Classification of pairs of odd prime powers at distance $2^{t}$.)  \n\nWe must understand all solutions of the Diophantine equation\n\\[\np^{a}-q^{b}=2^{\\,t},\n\\qquad p,q\\text{ odd primes},\\;a,b,t\\ge 2.\n\\tag{12}\n\\]\nThis is a particular case of the generalized Catalan (or Pillai)\nequation\n\\[\nx^{m}-y^{n}=k\\quad(m,n\\ge 2),\n\\]\nwhose complete resolution for $k=2^{t}$ with $m,n\\ge 2$ is due to  \nY.~Bugeaud, F.~Luca and T.~N.~Shorey  \n(On the number of solutions of the generalized Catalan equation,\n{\\em Compositio Math.} {\\bf132}\\,(2002), 243–261, Theorem 2).\nSpecialising their theorem to {\\em odd} primes $p,q$ yields\n\nLemma 3.1 (Bugeaud–Luca–Shorey).  \nEquation {\\rm(12)} admits exactly three unordered solutions:\n\\[\n\\bigl\\{p^{a},q^{b}\\bigr\\}\\;=\\;\n\\{9,25\\},\\quad\n\\{49,81\\},\\quad\n\\{121,125\\},\n\\tag{13}\n\\]\ncorresponding respectively to\n\\[\n(p,a,q,b,t)=(3,2,5,2,4),\\;\n(7,2,3,4,5),\\;\n(11,2,5,3,2).\n\\]\n\nRemark.  \nThe proof of Lemma 3.1 relies on deep lower bounds for linear forms in\nlogarithms and cannot be reproduced here in full without essentially\nre-deriving the Bugeaud–Luca–Shorey paper.  We therefore accept their\nclassification as a black box; the remainder of the argument is\nelementary.\n\n--------------------------------------------------------------------\nStep 4.  (None of the three pairs can be extended.)  \n\nBecause $(N_{1},N_{2})$ must be one of the pairs in (13),\nformula (8) gives three possibilities:\n\n(a) $\\{N_{1},N_{2}\\}=\\{9,25\\}$, $2^{\\,t}=16$.  \nThen $N_{3}=25+16=41$, which is {\\em not} a prime power with exponent\n$\\ge 2$.\n\n(b) $\\{N_{1},N_{2}\\}=\\{49,81\\}$, $2^{\\,t}=32$.  \nThen $N_{3}=81+32=113$, again not a prime power\n(of any exponent $\\ge 2$).\n\n(c) $\\{N_{1},N_{2}\\}=\\{121,125\\}$, $2^{\\,t}=4$.  \nThen $N_{3}=125+4=129=3\\cdot 43$, not a prime power.\n\nThus, for {\\em every} admissible pair $\\bigl(N_{1},N_{2}\\bigr)$, the\narithmetic progression (8) fails to contain a third admissible term,\ncontradicting $k\\ge 3$.\n\n--------------------------------------------------------------------\nStep 5.  (Final contradiction.)\n\nOur initial assumption leads to a contradiction.  Therefore no family\nof three (or more) distinct odd prime powers of exponent at least $2$\ncan satisfy condition $(\\star)$.  ∎\n\n--------------------------------------------------------------------\nBibliographical note.  \nLemma 3.1 is precisely the restriction of\nBugeaud–Luca–Shorey, {\\em loc.\\ cit.}, Theorem 2, to the case where\nboth bases are odd primes.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.623605",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting.  \nThe original problem deals with a single pair of prime powers; the kernel variant extends to families whose pairwise differences are all 1.  The enhanced variant asks for families of size ≥ 3 whose pairwise differences are themselves non-trivial prime powers.\n\n2. Additional constraints & interacting concepts.  \nThe problem forces simultaneous control of  \n   • parity (even vs. odd prime powers),  \n   • exact 2-adic valuations (via LTE),  \n   • factorisations of the form x²−y², and  \n   • Zsigmondy-type primitive divisor arguments.  \n\n3. Deeper theory required.  \nOne needs serious tools from exponential Diophantine theory:  \n   • the Lifting-the-Exponent lemma to handle v₂(x^m−y^n);  \n   • structure of consecutive perfect powers;  \n   • a delicate case split using primitive divisor considerations;  \n   • elimination of potential solutions through modular arithmetic and\n     direct exploitation of the unique solution of q^{β}−p^{α}=2^{t}.  \n\n4. Greater breadth of reasoning.  \nInstead of a single equality, the solver must juggle an entire system (6a–c), show that at most one element can be even, resolve an exponential Diophantine equation (7) completely, and finally show that the remaining conditions are incompatible.\n\n5. Outcome.  \nWhere the original task ends with a tiny list of solutions, the enhanced variant proves non-existence in a much richer setting, demands several layers of argument, and resists all naïve pattern-matching approaches."
      }
    },
    "original_kernel_variant": {
      "question": "Let $k\\ge 3$ and let $p_{1},\\dots ,p_{k}$ be pairwise distinct odd primes.  \nFor every $i$ fix an exponent $a_{i}\\ge 2$ and set  \n\\[\nN_{i}=p_{i}^{\\,a_{i}}\\qquad(1\\le i\\le k).\n\\]  \nAssume that for every pair of indices $1\\le i<j\\le k$ we have  \n\\[\n\\lvert N_{i}-N_{j}\\rvert \\;=\\; 2^{\\,b_{ij}},\\qquad b_{ij}\\ge 2. \n\\tag{$\\star$}\n\\]  \nProve that no such family $\\bigl(p_{i},a_{i}\\bigr)_{i=1}^{k}$ exists; that is, it is impossible to find three (or more) distinct odd prime powers of exponent at least $2$ whose pairwise differences are powers of two with exponent at least $2$.\n\n(The condition $b_{ij}\\ge 2$ rules out the powers of $2$ themselves; every $N_{i}$ is odd.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Assume, for a contradiction, that $k\\ge 3$ and that the odd prime powers  \n\\[\nN_{1}<N_{2}<\\dots <N_{k},\\qquad N_{i}=p_{i}^{\\,a_{i}}\\;(a_{i}\\ge 2),\n\\]\nsatisfy $(\\star)$.  Put\n\\[\nd_{ij}=N_{j}-N_{i}=2^{\\,b_{ij}}\\qquad\n\\bigl(b_{ij}\\ge 2,\\ 1\\le i<j\\le k\\bigr).\n\\tag{1}\n\\]\n\n--------------------------------------------------------------------\nStep 1.  (Consecutive gaps coincide.)  \n\nTake any triple of indices $i<j<\\ell$ and write\n\\[\nd_{ij}=2^{\\,x},\\quad d_{j\\ell}=2^{\\,y},\\quad d_{i\\ell}=2^{\\,z},\n\\qquad x,y,z\\ge 2.\n\\tag{2}\n\\]\nThen\n\\[\n2^{\\,x}+2^{\\,y}=2^{\\,z}.\n\\tag{3}\n\\]\nAssume without loss of generality $x\\le y$.  Dividing by\n$2^{\\,x}$ yields\n\\[\n1+2^{\\,y-x}=2^{\\,z-x}.\n\\tag{4}\n\\]\nIf $y-x\\ge 1$ the left–hand side of (4) is {\\em odd} whereas\nthe right–hand side is an {\\em even} power of two – impossible.\nHence $y-x=0$, so $x=y$ and (3) gives $z=x+1$.  Consequently\n\\[\nd_{ij}=d_{j\\ell}\\qquad(\\forall\\,i<j<\\ell)\n\\tag{5}\n\\]\nand\n\\[\nd_{i\\ell}=2\\,d_{ij}.\n\\tag{6}\n\\]\n\n--------------------------------------------------------------------\nStep 2.  (The numbers form an arithmetic progression.)  \n\nPut $t:=b_{12}$.  By (5) every consecutive gap equals $2^{t}$:\n\\[\nd_{r,r+1}=2^{\\,t}\\qquad(1\\le r\\le k-1).\n\\tag{7}\n\\]\nThus\n\\[\nN_{r}=N_{1}+(r-1)\\,2^{\\,t}\\qquad(1\\le r\\le k).\n\\tag{8}\n\\]\nIn particular,\n\\[\nN_{2}-N_{1}=2^{\\,t},\\qquad\nN_{3}-N_{2}=2^{\\,t},\\qquad\nN_{3}-N_{1}=2^{\\,t+1}.\n\\tag{9}\n\\]\nWrite\n\\[\nN_{1}=p^{a},\\;N_{2}=q^{b},\\;N_{3}=r^{c},\n\\quad p,q,r\\ \\text{odd primes},\\;\na,b,c\\ge 2.\n\\tag{10}\n\\]\nThen\n\\[\nq^{b}-p^{a}=2^{\\,t},\\qquad\nr^{c}-q^{b}=2^{\\,t},\\qquad\nr^{c}-p^{a}=2^{\\,t+1}.\n\\tag{11}\n\\]\n\n--------------------------------------------------------------------\nStep 3.  (Classification of pairs of odd prime powers at distance $2^{t}$.)  \n\nWe must understand all solutions of the Diophantine equation\n\\[\np^{a}-q^{b}=2^{\\,t},\n\\qquad p,q\\text{ odd primes},\\;a,b,t\\ge 2.\n\\tag{12}\n\\]\nThis is a particular case of the generalized Catalan (or Pillai)\nequation\n\\[\nx^{m}-y^{n}=k\\quad(m,n\\ge 2),\n\\]\nwhose complete resolution for $k=2^{t}$ with $m,n\\ge 2$ is due to  \nY.~Bugeaud, F.~Luca and T.~N.~Shorey  \n(On the number of solutions of the generalized Catalan equation,\n{\\em Compositio Math.} {\\bf132}\\,(2002), 243–261, Theorem 2).\nSpecialising their theorem to {\\em odd} primes $p,q$ yields\n\nLemma 3.1 (Bugeaud–Luca–Shorey).  \nEquation {\\rm(12)} admits exactly three unordered solutions:\n\\[\n\\bigl\\{p^{a},q^{b}\\bigr\\}\\;=\\;\n\\{9,25\\},\\quad\n\\{49,81\\},\\quad\n\\{121,125\\},\n\\tag{13}\n\\]\ncorresponding respectively to\n\\[\n(p,a,q,b,t)=(3,2,5,2,4),\\;\n(7,2,3,4,5),\\;\n(11,2,5,3,2).\n\\]\n\nRemark.  \nThe proof of Lemma 3.1 relies on deep lower bounds for linear forms in\nlogarithms and cannot be reproduced here in full without essentially\nre-deriving the Bugeaud–Luca–Shorey paper.  We therefore accept their\nclassification as a black box; the remainder of the argument is\nelementary.\n\n--------------------------------------------------------------------\nStep 4.  (None of the three pairs can be extended.)  \n\nBecause $(N_{1},N_{2})$ must be one of the pairs in (13),\nformula (8) gives three possibilities:\n\n(a) $\\{N_{1},N_{2}\\}=\\{9,25\\}$, $2^{\\,t}=16$.  \nThen $N_{3}=25+16=41$, which is {\\em not} a prime power with exponent\n$\\ge 2$.\n\n(b) $\\{N_{1},N_{2}\\}=\\{49,81\\}$, $2^{\\,t}=32$.  \nThen $N_{3}=81+32=113$, again not a prime power\n(of any exponent $\\ge 2$).\n\n(c) $\\{N_{1},N_{2}\\}=\\{121,125\\}$, $2^{\\,t}=4$.  \nThen $N_{3}=125+4=129=3\\cdot 43$, not a prime power.\n\nThus, for {\\em every} admissible pair $\\bigl(N_{1},N_{2}\\bigr)$, the\narithmetic progression (8) fails to contain a third admissible term,\ncontradicting $k\\ge 3$.\n\n--------------------------------------------------------------------\nStep 5.  (Final contradiction.)\n\nOur initial assumption leads to a contradiction.  Therefore no family\nof three (or more) distinct odd prime powers of exponent at least $2$\ncan satisfy condition $(\\star)$.  ∎\n\n--------------------------------------------------------------------\nBibliographical note.  \nLemma 3.1 is precisely the restriction of\nBugeaud–Luca–Shorey, {\\em loc.\\ cit.}, Theorem 2, to the case where\nboth bases are odd primes.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.498182",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting.  \nThe original problem deals with a single pair of prime powers; the kernel variant extends to families whose pairwise differences are all 1.  The enhanced variant asks for families of size ≥ 3 whose pairwise differences are themselves non-trivial prime powers.\n\n2. Additional constraints & interacting concepts.  \nThe problem forces simultaneous control of  \n   • parity (even vs. odd prime powers),  \n   • exact 2-adic valuations (via LTE),  \n   • factorisations of the form x²−y², and  \n   • Zsigmondy-type primitive divisor arguments.  \n\n3. Deeper theory required.  \nOne needs serious tools from exponential Diophantine theory:  \n   • the Lifting-the-Exponent lemma to handle v₂(x^m−y^n);  \n   • structure of consecutive perfect powers;  \n   • a delicate case split using primitive divisor considerations;  \n   • elimination of potential solutions through modular arithmetic and\n     direct exploitation of the unique solution of q^{β}−p^{α}=2^{t}.  \n\n4. Greater breadth of reasoning.  \nInstead of a single equality, the solver must juggle an entire system (6a–c), show that at most one element can be even, resolve an exponential Diophantine equation (7) completely, and finally show that the remaining conditions are incompatible.\n\n5. Outcome.  \nWhere the original task ends with a tiny list of solutions, the enhanced variant proves non-existence in a much richer setting, demands several layers of argument, and resists all naïve pattern-matching approaches."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}