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

(→‎Solution 2)
(→‎Solution 2)
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We can now see that <math>|x| \ge 2</math>. Furthermore, notice that if
 
We can now see that <math>|x| \ge 2</math>. Furthermore, notice that if
  
−
<math>x+1 = 3^k7^j</math>
+
<math>x+1 = \pm3^k7^j</math>
  
 
for a pair of positive integers <math>(j,k)</math> means that  
 
for a pair of positive integers <math>(j,k)</math> means that  
  
−
<math>x^2-x+1 \le 3</math>
+
<math>|x^2-x+1| \le 3</math>
  
 
which cannot be true. We now know that  
 
which cannot be true. We now know that  
  
−
<math>x+1 = 3^k, 7^k \rightarrow x^2-x+1 = 3*7^m, 3^m</math>.
+
<math>x+1 = \pm3^k, \pm7^k \rightarrow x^2-x+1 = 3*7^m, 3^m</math>.
  
 
Suppose that  
 
Suppose that  
  
−
<math>x+1 = 7^k \rightarrow (x+1)^2-3(x+1)+3 = 3^m \rightarrow 7^{2k}-3*7^k+3  = 3^m \rightarrow 7^{2k} \equiv 0\pmod{3}</math>  
+
<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  
 
which is a contradiction. Now suppose that  
  
−
<math>x+1 = 3^k \rightarrow (x+1)^2-3(x+1)+3 = 3*7^m \rightarrow 3^{2k}-3^{k+1}+3 = 3*7^m \rightarrow 3^{2k-1}-3^{k}+1 = 7^m \rightarrow 3^k(3^{k-1}-1) = 7^m-1</math>.
+
<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
 
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 = 1+v_3(m) \le 1+log_3(m) \rightarrow 7^m-1 = 3^k(3^{k-1}-1)\le 3^{1+log_3(m)}(3^{log_3(m)}-1) = 3m(m-1)</math>.
+
<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 > 0</math> we can see that  
+
For <math>m > 2</math> we can see that  
  
−
<math>7^m-1 > 3m(m-1)</math>  
+
<math>7^m-1 > 9m(3m\pm1)</math>  
  
 
which is a contradiction. Therefore there only possible solution is when  
 
which is a contradiction. Therefore there only possible solution is when  
  
−
<math>m = 0 \rightarrow x = 2</math>.
+
<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
 
Plugging this back into (1) and (2) yields
  
−
(4) <math>y+8 = 7^{314}3^{155} \rightarrow (7^{314}3^{155})(7^{314}3^{155}-7) = (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)</math>.
+
(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.
 
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.

Revision as of 13:15, 24 March 2012

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 $z$.

Solution

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, then 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. $\blacksquare$

See also

  • <url>Forum/viewtopic.php?p=213009#213009 Discussion on AoPS/MathLinks</url>
2005 USAMO (Problems • Resources)
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