Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 USAMO Problems/Problem 2"

(→‎Solution)
m
Line 7: Line 7:
 
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>.
  
−
==Hint==
 
 
Consider solving the problem by taking both equations '''mod 19'''.
 
Consider solving the problem by taking both equations '''mod 19'''.
  
−
 
+
== Solutions ==
−
 
+
=== Solution 1 ===
−
 
 
−
== Solution ==
 
 
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 36: Line 33:
 
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 ==  
+
=== Solution 2 ===
  
 
Note that the given can be rewritten as  
 
Note that the given can be rewritten as  
Line 133: Line 130:
  
 
Therefore there are no solutions to the given system of diophantine equations. <math>\blacksquare</math>
 
Therefore there are no solutions to the given system of diophantine equations. <math>\blacksquare</math>
 +
 +
{{alternate solutions}}
  
 
== See also ==
 
== See also ==

Revision as of 16:13, 19 July 2014

Problem

(Răzvan Gelca) Prove that the system \begin{align*}x^6 + x^3 + x^3y + y &= 147^{157} \\ x^3 + x^3y + y^2 + y + z^9 &= 157^{147}\end{align*} has no solutions in integers $x$, $y$, and $z$.

Consider solving the problem by taking both equations mod 19.

Solutions

Solution 1

It suffices to show that there are no solutions to this system in the integers mod 19. We note that $152 = 8 \cdot 19$, so $157 \equiv -147 \equiv 5 \pmod{19}$. For reference, we construct a table of powers of five: \[\begin{array}{c|c||c|c} n& 5^n &n & 5^n \\\hline 1 & 5 & 6 & 7 \\ 2 & 6 & 7 & -3 \\ 3 & -8 & 8 & 4 \\ 4 & -2 & 9 & 1 \\ 5 & 9 && \end{array}\] 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 $147^{157} \equiv (-5)^{13} \equiv -5^4 \equiv 2$, and $157^{147} \equiv 5^3 \equiv -8$. Thus we rewrite our system thus: \begin{align*} (x^3+y)(x^3+1) &\equiv 2 \\ (x^3+y)(y+1) + z^9 &\equiv -8 . \end{align*} Adding these, we have 

\[(x^3+y+1)^2 - 1 + z^9 &\equiv -6,\] (Error compiling LaTeX. Unknown error_msg)

or \[(x^3+y+1)^2 \equiv -z^9 - 5 .\] By Fermat's Little Theorem, the only possible values of $z^9$ are $\pm 1$ and 0, so the only possible values of $(x^3+y+1)^2$ are $-4,-5$, and $-6$. 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. $\blacksquare$

Solution 2

Note that the given can be rewritten as

(1) $(x^3+y)(x^3+1) = (x^3+y)(x+1)(x^2-x+1) = 147^{157} = 7^{314}3^{152}$,

(2) $(x^3+y)(y+1)+z^9 = 157^{147} \rightarrow (x^3+y)(y+1) = (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)$.

We can also see that

$x^2-x+1 = (x+1)^2-3(x+1)+3 \rightarrow \gcd(x+1, x^2-x+1) \le 3$.

Now we notice

$x^3+1 = 3^a7^b$

for some pair of non-negative integers $(a,b)$. We also note that

$x^2 \ge 2x$ when $|x| \ge 2 \rightarrow x^2-x+1 \ge x+1$

when $|x| \ge 2$. If $x = 1$ or $x = -1$ then examining (1) would yield $147^{157} \equiv 0 \pmod{2}$ which is a contradiction. If $x = 0$ then from (1) we can see that $y+1 = 147^{157}$, plugging this into 2 yields

(3) $(147^{157}-1)(147^{157}) = (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)$, $146 | (147^{157}-1)$, $146 = 2*73$.

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.

$73 | (157^{49}-z^3) \rightarrow z^3 \equiv 11^{49} \pmod{73} \rightarrow (11^{49})^{24} \equiv 1 \pmod{73}$.

However

$(11^{49})^{24} \equiv 8 \pmod{73}$

Thus we can see that 73 cannot divide the first factor in the right hand side of (3). Let us consider the next case.

$73 | (157^{98}+157^{49}z^3+z^6) \rightarrow 11^{98}+11^{49}z^3+z^6 \equiv 0 \pmod{73}$.

However

$11^{98}+11^{49}z^3+z^6 \equiv z^6+47z^3+19 \equiv z^6-26z^3+19 \equiv (z^3-13)^2-150 \equiv (z^3-13)^2-4 \equiv (z^3-11)(z^3-15) \pmod{73}$.

It can be seen that 11 and 15 are not perfect cubes $\pmod{73}$ from the following.

$15^{24} \equiv 11^{24} \equiv 8 \not\equiv 1 \pmod{73}$

We can now see that $|x| \ge 2$. Furthermore, notice that if

$x+1 = \pm3^k7^j$

for a pair of positive integers $(j,k)$ means that

$|x^2-x+1| \le 3$

which cannot be true. We now know that

$x+1 = \pm3^k, \pm7^k \rightarrow x^2-x+1 = 3*7^m, 3^m$.

Suppose that

$x+1 = \pm7^k \rightarrow (x+1)^2-3(x+1)+3 = 3^m \rightarrow 7^{2k}\pm3*7^k+3 = 3^m \rightarrow 7^{2k} \equiv 0\pmod{3}$

which is a contradiction. Now suppose that

$x+1 = \pm3^k \rightarrow (x+1)^2-3(x+1)+3 = 3*7^m \rightarrow 3^{2k}\pm3^{k+1}+3 = 3*7^m \rightarrow 3^{2k-1}\pm3^{k}+1 = 7^m \rightarrow 3^k(3^{k-1}\pm1) = 7^m-1$.

We now apply the lifting the exponent lemma to examine the power of 3 that divides each side of the equation when $m > 0$ to obtain

$k \le 2+v_3(m) \le 2+log_3(m) \rightarrow 7^m-1 = 3^k(3^{k-1}\pm1)\le 3^{2+log_3(m)}(3^{1+log_3(m)}\pm1) = 9m(3m\pm1)$.

For $m > 2$ we can see that

$7^m-1 > 9m(3m\pm1)$

which is a contradiction. Therefore there only possible solution is when

$m = 0 \rightarrow k = 1 \rightarrow x = 2, m = 1 \rightarrow k = 1 \rightarrow x = -4, m = 2 \rightarrow$ no integer solutions for k.

Plugging this back into (1) and (2) yields

(4) $3^{155} | (x^3+y) \rightarrow 3^{155}| (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)$.

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.

$9 | 157^{49}-z^3 \rightarrow 157^{49}-z^3 \equiv 0 \pmod{9}$.

However,

$z^3 \equiv 0, 1, 8\pmod{9}, 157^{49}-z^3 \equiv 4-z^3 \not\equiv 0 \pmod{9}$.

We now consider the second case.

$9 | (z^6+157^{49}z^3+157^{98}) \rightarrow 157^{98}+157^{49}z^3+z^6\equiv 0\pmod{9}$.

However

$z^6+157^{49}z^3+157^{98} \equiv z^6+4z^3+7 \equiv (z^3+2)^2+3 \not\equiv 0\pmod{9}$

Therefore there are no solutions to the given system of diophantine equations. $\blacksquare$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_2&oldid=62548"