antique math problem

This is a math problem published in a periodical in 1877.

Find the least integral values of x and y that will satisfy the equation:

x^2 - 9781*y^2 = 1.

I thought the least integral roots are (1, 0) and (-1, 0).

Also I noticed the possible x should be around 99*y, so I manually caculated about 50 pairs, no luck.

Then I wrote a program to look through the vicinity at (99*y - 10 < x < 99*y + 10, y) with y values going up to 10^9. I have a few lines of crap print out  because the computer was messed up when the number is too big.  Like these:

The roots are ---
 x = -1, y = 0
The roots are ---
 x = 1, y = 0
The roots are ---
 x = -1177971319, y = 161635332
The roots are ---
 x = 2125380257, y = 542070584
The roots are ---
 x = 879440713, y = 572868844
The roots are ---
 x = -1134323873, y = 856212344

Then I thought at that time (1877), pleople just did not have computers. Then I try to prove it.

Then I spent about 20 minutes to prove that there is no other integral roots using some dumb basic methods. By checking last digit of 9781*y^2 + 1, I rule out 9, 1, 4, 6 as the last digit of y. If the last digit of y is 7 or 3, then 9781*y^2 + 1 will end with 10, 30, 50, 70 or 90. If the last digit of y is 5, then last 2 digits of 9781*y^2 +1 are 26. No integer has square value ending in 26. If the last digit of y is 8 or 2, then y must be form of 50k + 2 or 50k -2, and x will take form 10m+5, then we will have  m(m+1)=9781k*(25k+1), which cannot be true. If the last digit of y is 0, then x must take form of 50k +1, and y is take the form of 10p, we will have 2500k^2+100k +1 = 978100m^2. This cannot be true. So this Diophantine equation has only roots (1, 0) and (-1, 0).

Can any one have a better solution to prove this? Thanks.

 

所有跟帖： 

• Thanks for your Q. I can do. -jinjing- ♀ (329 bytes) () 01/26/2011 postreply 08:40:22

• For srict logic,though it is trivial, | y^2--9781|=2 no intg. so -jinjing- ♀ (0 bytes) () 01/26/2011 postreply 10:28:12

• 回复：Thanks for your Q. I can do. -calligraphy- ♂ (681 bytes) () 01/26/2011 postreply 12:20:42

• We only check |y^2-P|=2,If there is no solution,the Q has no who -jinjing- ♀ (298 bytes) () 01/26/2011 postreply 16:50:24

• (n^2+1)^2-(n^2+2)*n^2=1,(n^2-1)^2-(n^2-2)*n^2=1.If for any coeff -jinjing- ♀ (229 bytes) () 01/26/2011 postreply 19:08:09

• 回复：(n^2+1)^2-(n^2+2)*n^2=1,(n^2-1)^2-(n^2-2)*n^2=1.If for any co -calligraphy- ♂ (511 bytes) () 01/26/2011 postreply 20:47:19

• Your Ex,5should be 5^2m,3 ...3^2n...,We get |9781-Z^2|=1, -jinjing- ♀ (344 bytes) () 01/27/2011 postreply 09:35:02

• 回复：Your Ex,5should be 5^2m,3 ...3^2n...,We get |9781-Z^2|=1, -calligraphy- ♂ (425 bytes) () 01/27/2011 postreply 11:26:44

• 回复：回复：Your Ex,5should be 5^2m,3 ...3^2n...,We get |9781-Z^2|=1, -jinjing- ♀ (438 bytes) () 01/27/2011 postreply 17:19:24

• 回复：We only check |y^2-P|=2,If there is no solution,the Q has no -calligraphy- ♂ (162 bytes) () 01/27/2011 postreply 07:03:34

• 回复：We only check |y^2-P|=2,If there is no solution,the Q has no -calligraphy- ♂ (3010 bytes) () 01/27/2011 postreply 07:11:39

• 回复：We only check |y^2-P|=2,If there is no solution,the Q has no -calligraphy- ♂ (380 bytes) () 01/27/2011 postreply 07:48:25

• 回复：We only check |y^2-P|=2,If there is no solution,the Q has no -calligraphy- ♂ (201 bytes) () 01/27/2011 postreply 08:21:54

• "So,Pi1=2,The right >=4. Contradiction, no solution if y>0",It i -jinjing- ♀ (0 bytes) () 01/28/2011 postreply 09:09:19

• 回复：antique math problem -calligraphy- ♂ (77 bytes) () 01/26/2011 postreply 13:46:05