diophantine equationsmodular arithmeticnumber theory
Working independently on some practice problems in preparation for my class exam and got stuck on the following question:
Show there are no solutions in integers $x$, $y$, and $z$ to the diophantine equation $x^2 + 2y^2 = 8z + 5.$
I know that we have to prove it using modular arithmetic but I'm not sure how to go about solving this problem. Could anyone help? Thanks!
The "topograph" for $x^2 - 13 y^2$ is definitely more complicated than the previous ones, because the continued fraction for $\sqrt {13}$ has period 5, your two previous examples had period 1. Confirming the "automorph" matrix, which just preserves the quadratic form:
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gp-pari
?
?
? form = [ 1,0; 0,-13]
%1 =
[1 0]
[0 -13]
?
? a = [649, 2340; 180, 649]
%2 =
[649 2340]
[180 649]
?
? atranspose = mattranspose(a)
%3 =
[649 180]
[2340 649]
?
? atranspose * form * a
%5 =
[1 0]
[0 -13]
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The pairs of numbers in green are vectors in the plane. Two basic properties. First, each shows its value for $x^2 - 13 y^2.$ For example, in the first occurrence of 4, we see the (column) vector $(11,3),$ and we can easily confirm that $11^2 - 13 \cdot 3^2 = 4. $ Next, around any point where three purple line segments meet (even if two are parallel), one of the three green vectors is the sum of the other two. For example, $$ (4,1) + (7,2) = (11,3). $$ As long as we just continue to the right, we can continue getting all positive entries in green.
Oh: you said you can do continued fractions. It happens that you can find all representations of 4 and 1 using the continued fraction of $\sqrt {13},$ so you can confirm a good deal of the Conway diagram, the vectors in green, whatever.
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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2
1 0 -13
0 form 1 0 -13 delta 0
1 form -13 0 1 delta 3
2 form 1 6 -4
-1 -3
0 -1
To Return
-1 3
0 -1
0 form 1 6 -4 delta -1
1 form -4 2 3 delta 1
2 form 3 4 -3 delta -1
3 form -3 2 4 delta 1
4 form 4 6 -1 delta -6
5 form -1 6 4 delta 1
6 form 4 2 -3 delta -1
7 form -3 4 3 delta 1
8 form 3 2 -4 delta -1
9 form -4 6 1 delta 6
10 form 1 6 -4
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$$5^{x} - y^2 = 4\tag{1}$$
We take the three cases $x=3a, x=3a+1$, and $x=3a+2.$
The problem can be reduced to finding the integer points on elliptic curves as follows.
$\bullet\ x=3a$
Let $X=5^{a}$, then we get
$y^2 =X^3 - 4.$
According to LMFDB, this elliptic curve has integral solutions $(X,y)=(2,\pm 2), (5,\pm 11).$
From $(5,\pm 11),$ we get $(x,y)=(3,11).$
$\bullet\ x=3a+1$
Let $X=5\cdot5^{a}, Y=5y$, then we get
$Y^2 =X^3 - 100.$
This elliptic curve has integral solutions $(X,Y)=(5,\pm 5),(10,\pm 30),(34,\pm 198).$
From $(5,\pm 5)$, we get $(x,y)=(1,1).$
$\bullet\ x=3a+2$
Let $X=25\cdot5^{a}, Y=25y$, then we get
$Y^2 =X^3 - 2500.$
This elliptic curve has integral solution $(X,Y)=(50,\pm 350).$
We get no solution $(x,y).$
Hence there are only integral solutions $(x,y)=(1,1),(3,11).$
Best Answer
If you look at the equation mod $8$, note that squares are either $0,1$ or $4$ mod 8. Thus the left hand side can only attain the values $0,1,2,3,4$ or $6$ mod $8$ and hence will never be equal to the right hand side, which is $5$ mod $8$.