Difference between revisions of "2005 USAMO Problems/Problem 2"
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has no solutions in integers <math>x</math>, <math>y</math>, and <math>z</math>. | has no solutions in integers <math>x</math>, <math>y</math>, and <math>z</math>. | ||
− | == Solution == | + | == Solutions == |
+ | === Solution 1 === | ||
It suffices to show that there are no solutions to this system in the integers mod 19. We note that <math>152 = 8 \cdot 19</math>, so <math>157 \equiv -147 \equiv 5 \pmod{19}</math>. For reference, we construct a table of powers of five: | It suffices to show that there are no solutions to this system in the integers mod 19. We note that <math>152 = 8 \cdot 19</math>, so <math>157 \equiv -147 \equiv 5 \pmod{19}</math>. For reference, we construct a table of powers of five: | ||
<cmath> \begin{array}{c|c||c|c} | <cmath> \begin{array}{c|c||c|c} | ||
Line 17: | Line 18: | ||
5 & 9 && | 5 & 9 && | ||
\end{array} </cmath> | \end{array} </cmath> | ||
− | Evidently, | + | Evidently, the order of 5 is 9. Hence 5 is the square of a multiplicative generator of the nonzero integers mod 19, so this table shows all nonzero squares mod 19, as well. |
It follows that <math>147^{157} \equiv (-5)^{13} \equiv -5^4 \equiv 2</math>, and <math>157^{147} \equiv 5^3 \equiv -8</math>. Thus we rewrite our system thus: | It follows that <math>147^{157} \equiv (-5)^{13} \equiv -5^4 \equiv 2</math>, and <math>157^{147} \equiv 5^3 \equiv -8</math>. Thus we rewrite our system thus: | ||
Line 25: | Line 26: | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
Adding these, we have | Adding these, we have | ||
− | <cmath> (x^3+y+1)^2 - 1 + z^9 | + | <cmath> (x^3+y+1)^2 - 1 + z^9 \equiv -6, </cmath> |
or | or | ||
<cmath> (x^3+y+1)^2 \equiv -z^9 - 5 . </cmath> | <cmath> (x^3+y+1)^2 \equiv -z^9 - 5 . </cmath> | ||
By [[Fermat's Little Theorem]], the only possible values of <math>z^9</math> are <math>\pm 1</math> and 0, so the only possible values of <math>(x^3+y+1)^2</math> are <math>-4,-5</math>, and <math>-6</math>. But none of these are squares mod 19, a contradiction. Therefore the system has no solutions in the integers mod 19. Therefore the solution has no equation in the integers. <math>\blacksquare</math> | By [[Fermat's Little Theorem]], the only possible values of <math>z^9</math> are <math>\pm 1</math> and 0, so the only possible values of <math>(x^3+y+1)^2</math> are <math>-4,-5</math>, and <math>-6</math>. But none of these are squares mod 19, a contradiction. Therefore the system has no solutions in the integers mod 19. Therefore the solution has no equation in the integers. <math>\blacksquare</math> | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | Note that the given can be rewritten as | ||
+ | |||
+ | (1) <math>(x^3+y)(x^3+1) = (x^3+y)(x+1)(x^2-x+1) = 147^{157} = 7^{314}3^{152}</math>, | ||
+ | |||
+ | (2) <math>(x^3+y)(y+1)+z^9 = 157^{147} \rightarrow (x^3+y)(y+1) = (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)</math>. | ||
+ | |||
+ | We can also see that | ||
+ | |||
+ | <math>x^2-x+1 = (x+1)^2-3(x+1)+3 \rightarrow \gcd(x+1, x^2-x+1) \le 3</math>. | ||
+ | |||
+ | Now we notice | ||
+ | |||
+ | <math>x^3+1 = 3^a7^b</math> | ||
+ | |||
+ | for some pair of non-negative integers <math>(a,b)</math>. We also note that | ||
+ | |||
+ | <math>x^2 \ge 2x</math> when <math>|x| \ge 2 \rightarrow x^2-x+1 \ge x+1</math> | ||
+ | |||
+ | when <math>|x| \ge 2</math>. If <math>x = 1</math> or <math>x = -1</math> then examining (1) would yield <math>147^{157} \equiv 0 \pmod{2}</math> which is a contradiction. If <math>x = 0</math> then from (1) we can see that <math>y+1 = 147^{157}</math>, plugging this into 2 yields | ||
+ | |||
+ | (3) <math>(147^{157}-1)(147^{157}) = (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)</math>, <math>146 | (147^{157}-1)</math>, <math>146 = 2*73</math>. | ||
+ | |||
+ | Noting that 73 is a prime number we see that it must divide at least 1 of the 2 factors on the right hand side of 3. Let us consider both cases. | ||
+ | |||
+ | <math>73 | (157^{49}-z^3) \rightarrow z^3 \equiv 11^{49} \pmod{73} \rightarrow (11^{49})^{24} \equiv 1 \pmod{73}</math>. | ||
+ | |||
+ | However | ||
+ | |||
+ | <math>(11^{49})^{24} \equiv 8 \pmod{73}</math> | ||
+ | |||
+ | Thus we can see that 73 cannot divide the first factor in the right hand side of (3). Let us consider the next case. | ||
+ | |||
+ | <math>73 | (157^{98}+157^{49}z^3+z^6) \rightarrow 11^{98}+11^{49}z^3+z^6 \equiv 0 \pmod{73}</math>. | ||
+ | |||
+ | However | ||
+ | |||
+ | <math>11^{98}+11^{49}z^3+z^6 \equiv z^6+47z^3+19 \equiv z^6-26z^3+19 \equiv (z^3-13)^2-150 \equiv (z^3-13)^2-4 \equiv (z^3-11)(z^3-15) \pmod{73}</math>. | ||
+ | |||
+ | It can be seen that 11 and 15 are not perfect cubes <math>\pmod{73}</math> from the following. | ||
+ | |||
+ | <math>15^{24} \equiv 11^{24} \equiv 8 \not\equiv 1 \pmod{73}</math> | ||
+ | |||
+ | We can now see that <math>|x| \ge 2</math>. Furthermore, notice that if | ||
+ | |||
+ | <math>x+1 = \pm3^k7^j</math> | ||
+ | |||
+ | for a pair of positive integers <math>(j,k)</math> means that | ||
+ | |||
+ | <math>|x^2-x+1| \le 3</math> | ||
+ | |||
+ | which cannot be true. We now know that | ||
+ | |||
+ | <math>x+1 = \pm3^k, \pm7^k \rightarrow x^2-x+1 = 3*7^m, 3^m</math>. | ||
+ | |||
+ | Suppose that | ||
+ | |||
+ | <math>x+1 = \pm7^k \rightarrow (x+1)^2-3(x+1)+3 = 3^m \rightarrow 7^{2k}\pm3*7^k+3 = 3^m \rightarrow 7^{2k} \equiv 0\pmod{3}</math> | ||
+ | |||
+ | which is a contradiction. Now suppose that | ||
+ | |||
+ | <math>x+1 = \pm3^k \rightarrow (x+1)^2-3(x+1)+3 = 3*7^m \rightarrow 3^{2k}\pm3^{k+1}+3 = 3*7^m \rightarrow 3^{2k-1}\pm3^{k}+1 = 7^m \rightarrow 3^k(3^{k-1}\pm1) = 7^m-1</math>. | ||
+ | |||
+ | We now apply the lifting the exponent lemma to examine the power of 3 that divides each side of the equation when <math>m > 0</math> to obtain | ||
+ | |||
+ | <math>k \le 2+v_3(m) \le 2+log_3(m) \rightarrow 7^m-1 = 3^k(3^{k-1}\pm1)\le 3^{2+log_3(m)}(3^{1+log_3(m)}\pm1) = 9m(3m\pm1)</math>. | ||
+ | |||
+ | For <math>m > 2</math> we can see that | ||
+ | |||
+ | <math>7^m-1 > 9m(3m\pm1)</math> | ||
+ | |||
+ | which is a contradiction. Therefore there only possible solution is when | ||
+ | |||
+ | <math>m = 0 \rightarrow k = 1 \rightarrow x = 2, m = 1 \rightarrow k = 1 \rightarrow x = -4, m = 2 \rightarrow</math> no integer solutions for k. | ||
+ | |||
+ | Plugging this back into (1) and (2) yields | ||
+ | |||
+ | (4) <math>3^{155} | (x^3+y) \rightarrow 3^{155}| (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)</math>. | ||
+ | |||
+ | In order for (4) to be true we must have 9 dividing at least 1 of the factors on the right hand side of the equation. Let us consider both cases. | ||
+ | |||
+ | <math>9 | 157^{49}-z^3 \rightarrow 157^{49}-z^3 \equiv 0 \pmod{9}</math>. | ||
+ | |||
+ | However, | ||
+ | |||
+ | <math>z^3 \equiv 0, 1, 8\pmod{9}, 157^{49}-z^3 \equiv 4-z^3 \not\equiv 0 \pmod{9}</math>. | ||
+ | |||
+ | We now consider the second case. | ||
+ | |||
+ | <math>9 | (z^6+157^{49}z^3+157^{98}) \rightarrow 157^{98}+157^{49}z^3+z^6\equiv 0\pmod{9}</math>. | ||
+ | |||
+ | However | ||
+ | |||
+ | <math>z^6+157^{49}z^3+157^{98} \equiv z^6+4z^3+7 \equiv (z^3+2)^2+3 \not\equiv 0\pmod{9}</math> | ||
+ | |||
+ | Therefore there are no solutions to the given system of diophantine equations. <math>\blacksquare</math> | ||
+ | |||
+ | === Solution 3 === | ||
+ | We will show there is no solution to the system modulo 13. Add the two equations and add 1 to obtain | ||
+ | <cmath>(x^3 + y + 1)^2 + z^9 = 147^{157} + 157^{147} + 1.</cmath> | ||
+ | By Fermat's Theorem, <math>a^{12}\equiv 1\pmod{13}</math> when <math>a</math> is not a multiple of 13. Hence we compute <math>147^{157}\equiv 4^1\equiv 4\pmod{13}</math> and <math>157^{147}\equiv 1^3\equiv 1\pmod{13}</math>. Thus | ||
+ | <cmath>(x^3 + y + 1)^2 + z^9\equiv 6\pmod{13}.</cmath> | ||
+ | The cubes mod 13 are <math>0</math>, <math>\pm 1</math>, and <math>\pm 5</math>. Writing the first equation as | ||
+ | <cmath>(x^3 + 1)(x^3 + y)\equiv 4\pmod{13},</cmath> | ||
+ | we see that there is no solution in case <math>x^3\equiv -1\pmod{13}</math> and for <math>x^3</math> congruent to <math>0, 1, 5, -5\pmod{13}</math>, correspondingly <math>x^3 + y</math> must be congruent to <math>4, 2, 5, -1</math>. Hence | ||
+ | <cmath>(x^3 + y + 1)^2\equiv \text{12, 9, 10, or 0}\pmod{13}.</cmath> | ||
+ | Also <math>z^9</math> is a cube, hence <math>z^9</math> must be <math>\text{0, 1, 5, 8, or 12}\pmod{13}</math>. It is easy to check that <math>6\pmod{13}</math> is not obtained by adding one of <math>0, 9, 10, 12</math> to one of <math>0, 1, 5, 8, 12</math>. Hence the system has no solutions in integers. | ||
+ | |||
+ | ''Note'': This argument shows there is no solution even if <math>z^9</math> is replaced by <math>z^3</math>. | ||
{{alternate solutions}} | {{alternate solutions}} | ||
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[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 09:38, 12 August 2015
Problem
(Răzvan Gelca) Prove that the system has no solutions in integers , , and .
Solutions
Solution 1
It suffices to show that there are no solutions to this system in the integers mod 19. We note that , so . For reference, we construct a table of powers of five: Evidently, the order of 5 is 9. Hence 5 is the square of a multiplicative generator of the nonzero integers mod 19, so this table shows all nonzero squares mod 19, as well.
It follows that , and . Thus we rewrite our system thus: Adding these, we have or By Fermat's Little Theorem, the only possible values of are and 0, so the only possible values of are , and . But none of these are squares mod 19, a contradiction. Therefore the system has no solutions in the integers mod 19. Therefore the solution has no equation in the integers.
Solution 2
Note that the given can be rewritten as
(1) ,
(2) .
We can also see that
.
Now we notice
for some pair of non-negative integers . We also note that
when
when . If or then examining (1) would yield which is a contradiction. If then from (1) we can see that , plugging this into 2 yields
(3) , , .
Noting that 73 is a prime number we see that it must divide at least 1 of the 2 factors on the right hand side of 3. Let us consider both cases.
.
However
Thus we can see that 73 cannot divide the first factor in the right hand side of (3). Let us consider the next case.
.
However
.
It can be seen that 11 and 15 are not perfect cubes from the following.
We can now see that . Furthermore, notice that if
for a pair of positive integers means that
which cannot be true. We now know that
.
Suppose that
which is a contradiction. Now suppose that
.
We now apply the lifting the exponent lemma to examine the power of 3 that divides each side of the equation when to obtain
.
For we can see that
which is a contradiction. Therefore there only possible solution is when
no integer solutions for k.
Plugging this back into (1) and (2) yields
(4) .
In order for (4) to be true we must have 9 dividing at least 1 of the factors on the right hand side of the equation. Let us consider both cases.
.
However,
.
We now consider the second case.
.
However
Therefore there are no solutions to the given system of diophantine equations.
Solution 3
We will show there is no solution to the system modulo 13. Add the two equations and add 1 to obtain By Fermat's Theorem, when is not a multiple of 13. Hence we compute and . Thus The cubes mod 13 are , , and . Writing the first equation as we see that there is no solution in case and for congruent to , correspondingly must be congruent to . Hence Also is a cube, hence must be . It is easy to check that is not obtained by adding one of to one of . Hence the system has no solutions in integers.
Note: This argument shows there is no solution even if is replaced by .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
- <url>Forum/viewtopic.php?p=213009#213009 Discussion on AoPS/MathLinks</url>
2005 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.