Diophantine equation of power 2

Question

I want to solve the following equation: $x^2-17y^2=104$ where $x,y$ are integers. I don't know how to proceed here. I have tried with mod $13$ and mod $17$ but it doesn't work.

Thanks in advance!

Answer 1

Above equation shown below:

$x^2-17y^2=104$

Above has parametric solution:

$x=[(11k^2-34k+187)/(k^2-17)]$

$y=[(k^2-22k-17)/(k^2-17)]$

For, $k=0$ we get:

$(x,y)=(11,1)$

Answer 2

This really does show all solutions with the variables positive.

In the output below, we have a recursion, three term, but referring to every fourth line. We could write $$ w_{n+8} = 66 w_{n+4} - w_n, $$ for example $$ 32923 = 66 \cdot 499 - 11 \; . \; $$ In the same way $$ v_{n+8} = 66 v_{n+4} - v_n, $$ for example $$ 7985 = 66 \cdot 121 - 1 \; . \; $$

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:  
    33   136
    8   33
  Automorphism backwards:  
    33   -136
    -8   33

  33^2 - 17 8^2 = 1

 w^2 - 17 v^2 = 104 =  2^3 13

Mon Apr 29 18:19:58 PDT 2019

w:  11  v:  1  SEED   KEEP +- 
w:  23  v:  5  SEED   KEEP +- 
w:  79  v:  19  SEED   BACK ONE STEP  23 ,  -5
w:  227  v:  55  SEED   BACK ONE STEP  11 ,  -1
w:  499  v:  121
w:  1439  v:  349
w:  5191  v:  1259
w:  14971  v:  3631
w:  32923  v:  7985
w:  94951  v:  23029
w:  342527  v:  83075
w:  987859  v:  239591
w:  2172419  v:  526889
w:  6265327  v:  1519565
w:  22601591  v:  5481691
w:  65183723  v:  15809375
w:  143346731  v:  34766689

Mon Apr 29 18:22:04 PDT 2019

 w^2 - 17 v^2 = 104 =  2^3 13

Answer 3

COMMENT.- Modulo $17$ one has $x^2=2$ whose solutions in $\mathbb F_{17}$ are $x=6$ and $x=11$ so we have in $\mathbb Z$

$$x=17n+6\text{ and } x=17n+11$$ It follows respectively $$y^2=17n^2+12n-4\text{ and } y^2=17n^2+22n+1$$ Then we have the identities $$(17n+6)^2-17(17n^2+12n-4)=104$$ $$(17m+11)^2-17(17m^2+22m+1)=104$$ It remains to study when $17n^2+12n-4$ and $17m^2+22m+1$ can be squares.

Some solutions of the first, $17n^2+12n-4=y^2$ are $n=-5,-1,1,13$ which give respectively $y=19,1,5,55$ so $ x=\sqrt{17y^2+104}=79,11,23,227$. Thus one has, as examples, the four solutions $$(x,y)=(79,19),(11,1),(23,5),(227,55)$$ but there must be many more, infinitely many probably.

Answer 4

It is clear that $y=1$ and $x=11$ is a solution of $x^2-17y^2=104$.

As pointed out by others, so are $(x,y)=(23,5), (79,19),$ and $(227,55)$.

Once you have a solution $(x,y)$ to $x^2-17y^2=104$, another is $(33x+136y, 8x+33y)$.

That's because $33-17\times8^2=(33-8\sqrt{17})(33+8\sqrt{17})=1$,

so if $x^2-17y^2=(x-y\sqrt{17})(x+y\sqrt{17})=104$

then $(33-8\sqrt{17})(x-y\sqrt{17})(33+8\sqrt{17})(x+y\sqrt{17})=104=$

$[(33x+8\times17y)-(8x+33y)\sqrt{17})][(33x+8\times17y)+(8x+33y)\sqrt{17}].$

The above information gives all of the solutions listed by Will Jagy.