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| − | ==Problem==
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| − | Are there integers <math> a </math> and <math> b </math> such that <math> a^5b+3 </math> and <math> ab^5+3 </math> are both perfect cubes of integers?
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| − | ==Solution==
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| − | No, such integers do not exist. This shall be proven by contradiction, by showing that if <math>a^5b+3</math> is a perfect cube then <math>ab^5+3</math> cannot be.
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| − | Remark that perfect cubes are always congruent to <math>0</math>, <math>1</math>, or <math>-1</math> modulo <math>9</math>. Therefore, if <math>a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}</math>, then <math>a^5b\equiv 5,6,\text{ or }7\pmod{9}</math>.
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| − | If <math>a^5b\equiv 6\pmod 9</math>, then note that <math>3|b</math>. (This is because if <math>3|a</math> then <math>a^5b\equiv 0\pmod 9</math>.) Therefore <math>ab^5\equiv 0\pmod 9</math> and <math>ab^5+3\equiv 3\pmod 9</math>, contradiction.
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| − | Otherwise, either <math>a^5b\equiv 5\pmod 9</math> or <math>a^5b\equiv 7\pmod 9</math>. Note that since <math>a^6b^6</math> is a perfect sixth power, and since neither <math>a</math> nor <math>b</math> contains a factor of <math>3</math>, <math>a^6b^6\equiv 1\pmod 9</math>. If <math>a^5b\equiv 5\pmod 9</math>, then <cmath>a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.</cmath> Similarly, if <math>a^5b\equiv 7\pmod 9</math>, then <cmath>a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.</cmath> Therefore <math>ab^5+3\equiv 5,7\pmod 9</math>, contradiction.
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| − | Therefore no such integers exist.
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| − | Amkan2022
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| − | ==Solution 2==
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| − | We shall prove that such integers do not exist via contradiction.
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| − | Suppose that <math>a^5b + 3 = x^3</math> and <math>ab^5 + 3 = y^3</math> for integers x and y. Rearranging terms gives <math>a^5b = x^3 - 3</math> and <math>ab^5 = y^3 - 3</math>. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = <math>(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}</math> and b = <math>(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}</math>. Consider a prime p in the prime factorization of <math>x^3 - 3</math> and <math>y^3 - 3</math>. If it has power <math>r_1</math> in <math>x^3 - 3</math> and power <math>r_2</math> in <math>y^3 - 3</math>, then <math>5r_1</math> - <math>r_2</math> is a multiple of 24 and <math>5r_2</math> - <math>r_1</math> also is a multiple of 24.
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| − | Adding and subtracting the divisions gives that <math>r_1</math> - <math>r_2</math> divides 12. (actually, <math>r_1 - r_2</math> is a multiple of 4, as you can verify if <math>\{^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}</math>. So the rest of the proof is invalid.) Because <math>5r_1</math> - <math>r_2</math> also divides 12, <math>4r_1</math> divides 12 and thus <math>r_1</math> divides 3. Repeating this trick for all primes in <math>x^3 - 3</math>, we see that <math>x^3 - 3</math> is a perfect cube, say <math>q^3</math>. Then <math>x^3 - q^3 = 3,</math> and <math>(x-q)(x^2 + xq + q^2) = 3</math>, so that <math>x - q = 1</math> and <math>x^2 + xq + q^2 = 3</math>. Clearly, this system of equations has no integer solutions for <math>x</math> or <math>q</math>, a contradiction, hence completing the proof.
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| − | Therefore no such integers exist.
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| − | ==Solution 3==
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| − | Let <math>a^5b+3=x^3</math> and <math>ab^5+3=y^3</math>. Then, <math>a^5b=x^3-3</math>, <math>ab^5=y^3-3</math>, and <cmath>(ab)^6=(x^3-3)(y^3-3)</cmath>
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| − | Now take <math>\text{mod }9</math> (recall that perfect cubes <math>\equiv -1,0,1\pmod{9}</math> and perfect sixth powers <math>\equiv 0,1\pmod{9}</math>) on both sides. There are <math>3\times 3=9</math> cases to consider on what values <math>\text{mod }9</math> that <math>x^3</math> and <math>y^3</math> take. Checking these <math>9</math> cases, we see that only <math>x^3\equiv y^3\equiv 0\pmod{9}</math> or <math>x\equiv y\equiv 0\pmod{3}</math> yield a valid residue <math>\text{mod }9</math> (specifically, <math>(x^3-3)(y^3-3)\equiv 0\pmod{9}</math>). But this means that <math>3\mid ab</math>, so <math>729\mid (ab)^6</math> so <cmath>729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)</cmath> contradiction.
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| − | ==Solution 4==
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| − | If <math>a^5b+3</math> is a perfect cube, then <math>a^5b</math> can be one of <math>5,6,7 \pmod 9</math>, so <math>(a^5b)^5 = a^{25} b^5</math> can be one of <math>5^5 \equiv 2</math>, <math>6^5 \equiv 0</math>, or <math>7^5 \equiv 4 \pmod 9</math>. If <math>a</math> were divisible by <math>3</math>, we'd have <math>a^5 b \equiv 0 \pmod 9</math>, which we've ruled out. So <math>\gcd(a,9) = 1</math>, which means <math>a^6 \equiv 1 \pmod 9</math>, and therefore <math>a^{25} b^5 \equiv ab^5 \pmod 9</math>.
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| − | We've shown that <math>a b^5</math> can be one of <math>0, 2, 4 \pmod 9</math>, so <math>ab^5 + 3</math> can be one of <math>3, 5, 7 \pmod 9</math>. None of these are possibilities for a perfect cube, so if <math>a^5b+3</math> is a perfect cube, <math>ab^5+3</math> cannot be.
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| − | ==Solution 5==
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| − | As in previous solutions, notice <math>ab^5,a^5b \equiv 5,6,7 \pmod 9</math>. Now multiplying gives <math>a^6b^6</math>, which is only <math>0,1 \pmod 9</math>, so after testing all cases we find that <math>ab^5\equiv a^5b \equiv 6 \mod 9</math>. Then since <math>\phi (9) = 6</math>, <math>ab^5\equiv \frac{a}{b}\pmod 9</math> and <math>a^5b \equiv \frac{b}{a}\pmod 9</math> (Note that <math>a,b</math> cannot be <math>0\pmod 9</math>). Thus we find that the inverse of <math>6</math> is itself under modulo <math>9</math>, a contradiction.
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| − | ==Solution 6==
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| − | I claim there are no such a or b such that both expressions are cubes.
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| − | Assume to the contrary <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes.
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| − | '''Lemma 1''': If <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes, then <math>ab^5, a^5b \equiv 5,7 \pmod 9</math>
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| − | '''Proof''' Since cubes are congruent to any of <math>0, 1, -1 \pmod 9</math>, <math>ab^5,a^5b \equiv 5,6,7 \pmod 9</math>. But if <math>ab^5 \equiv 6 \pmod 9</math>, <math>3|a</math>, so <math>a^5b \equiv 0 \pmod 9</math>, contradiction. A similar argument can be made for <math>ab^5 \neq 6 \pmod 9</math>.
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| − | '''Lemma 2''': If k is a perfect 6th power, then <math>k \equiv 0,1 \pmod 9</math>
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| − | '''Proof''': Since cubes are congruent to <math>0, 1, -1 \pmod 9</math>, we can square, and get 6th powers are congruent to <math>0, 1 \pmod 9</math>.
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| − | Since <math>ab^5 \cdot a^5b = a^6 b^6 = (ab)^6</math>, which is a perfect 6th power, by lemma 2, <math>ab^5 \cdot a^5b \equiv 0,1 \pmod 9</math>.
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| − | But, by lemma 1, <math>ab^5 \cdot a^5b \equiv 5 \cdot 5, 5 \cdot 7, 7 \cdot 7 \equiv 7, 8, 4 \pmod 9</math>.
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| − | So, <math>ab^5 \cdot a^5b</math>, which is an integer, can't go into any of the possible residue classes modulo 9, without breaking one of these lemmas. This is a contradiction, and the proof is complete. <math>\blacksquare</math>
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| − | -AlexLikeMath
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| − | ==Solution 7==
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| − | Note that <math>x^3 \equiv 0,1,-1 \pmod 9</math>. So <math>a^5b+3, ab^5+3\equiv 0,1,-1\pmod{9}</math><math>\Rightarrow a^5b, ab^5 \equiv 5,6,7 \pmod 9</math>.
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| − | <math>a^5b \equiv 5 \pmod 9 \;\;\;\;\;\; ab^5 \equiv 5 \pmod 9</math>
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| − | <math>a^5b \equiv 6 \pmod 9 \;\;\;\;\;\; ab^5 \equiv 6 \pmod 9</math>
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| − | <math>a^5b \equiv 7 \pmod 9 \;\;\;\;\;\; ab^5 \equiv 7 \pmod 9</math>.
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| − | Multiplying <math>a^5b</math> and <math>ab^5</math> we get <math>(ab)^6</math>. By substituting values for <math>a^5b</math> and <math>ab^5</math> we can see that <math>(ab)^6 \equiv 0,1,3,6,7,8 \pmod 9</math> but because cubes are not <math>3,6,7 \pmod 9</math>, and squares are not <math>-1\pmod 9</math> so <math>\boxed{(ab)^6 \equiv 0,1 \pmod 9}</math>.
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| − | *Case I: <math>(ab)^6\equiv 0 \pmod 9</math>.This is only true when <math>3|ab</math>. If <math>3|a,b</math> then <math>a^5b+3 \equiv 3 \pmod 9</math>, which is not a perfect cube. The same can be seen with if <math>3|a</math> then <math>a^5b+3 \equiv 3 \pmod 9</math> and if <math>3|a</math> then <math>b^5a+3 \equiv 3 \pmod 9</math>. So 6 mod 9 case is eliminated from the above relations of congruences
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| − | *Case II: <math>(ab)^6\equiv 1 \pmod 9</math> for this we'll take a look at the above information regarding <math>a^5b</math> and <math>ab^5</math>. If <math>a^5b \equiv 5 \pmod 9</math> then,
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| − | <math>(ab)^6=a^5b \cdot ab^5 \equiv 5(ab^5) \pmod 9 \Rightarrow ab^5 \equiv 2 \pmod 9 \Rightarrow ab^5+3 \equiv
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| − | 5 \pmod 9</math>
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| − | which is not possible. Similarly, if <math>a^5b \equiv 7 \pmod 9</math> then,
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| − | <math>(ab)^6=a^5b \cdot ab^5 \equiv 7(ab^5) \pmod 9 \Rightarrow ab^5 \equiv 4 \pmod 9 \Rightarrow ab^5+3 \equiv
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| − | 7 \pmod 9</math>
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| − | which is not possible. So, no such integers.
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| − | -Lakshya Pamecha
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| − | ==Solution 8==
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| − | We proceed by contradiction. Assuming <math>a^5b + 3 = x^3</math> for some integer <math>x</math> and <math>ab^5 + 3 = y^3</math> for another integer <math>y</math>, we can multiply the two equations to get <math>a^6b^6 + 3a^5b + 3ab^5 + 9 = x^3y^3</math>. We can rearrange this to <math>3a^5b+3ab^5 = x^3y^3 - a^6b^6 +9</math>.
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| − | Looking at this <math>\pmod 9</math>, this becomes <math>3a^5b+3ab^5 \equiv x^3y^3-a^6b^6 \pmod 9</math>. Looking at the right hand side, the first term is either -1, 0, or 1 <math>\pmod 9</math>. The second is 0 or 1 <math>\pmod 9</math>, because it is a square of a cube. Additionally, since the LHS is divisible by three, so is the RHS.
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| − | The only achievable value for the LHS that fits this is 0. So, <math>3a^5b+3ab^5 \equiv 0 \pmod 9</math>. Dividing by three, making sure to divide the modulus as well, <math>a^5b + ab^5 \equiv 0 \pmod 3</math>. Noticing <math>a^5b \equiv ab \pmod 3</math> and <math>ab^5 \equiv ab \pmod 3</math>, we get, <math>2ab \equiv 0 \pmod 3</math>.
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| − | This means either <math>3|a</math> or <math>3|b</math>, or both. WLOG, <math>3|a</math>. Looking at the original expression mod 9, we get <math>a^5b + 3 \equiv -1, 0 </math>,or <math> 1 \pmod 9</math>. Since <math>a</math> is divisible by three, <math>a^5</math> is divisible by 243, so it is divisible by 9. This reduces the equation to <math>3 \equiv -1, 0</math>,or <math> 1 \pmod 9</math>, which obviously can't be true, a contradiction. Therefore, our original assumption, that both expressions were perfect cubes, is incorrect.
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| − | So, no, no such integers exist.
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| − | -sillybone(plz tell me if there's a flaw in the solution, I'm new to editing pages)
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| − | ==Note==
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| − | After you get that <math>a^5 b \equiv 5, 6, \text{or} 7 \pmod{9}</math>, you can also bash by listing all possible residues for <math>a^5</math> and their corresponding values for <math>b</math>. At that point, you can easily see that all the solutions for <math>a</math> and <math>b</math> do not work in the other equation, meaning that there are no solutions that satisfy both equations.
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| − | {{MAA Notice}}
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