2
$\begingroup$

I found out that in a primitive pythagorean triple $$a^2+b^2=c^2$$ the difference $d=|a-b|$ (which must be odd) can occur, if and only if we can write $$d=a^2-2b^2$$ with positive coprime integers $a,b$. Moreover, $d$ is a possible difference if and only if $-d$ is a possible difference. We can replace the pair $(a/b)$ by $(a+2b/a+b)$ to get a solution of the desired form.

When can an odd integer $d$ be written as $d=a^2-2b^2$ with positive coprime integers $a,b$ ?

The representation $49=9^2-2\cdot 4^2$ shows that $d$ need not be squarefree.

$\endgroup$
  • 1
    $\begingroup$ I know that this is a pell-type equation but I wonder how we can verify whether a coprime solution exists, this might be harder than to verify whether a solution exists at all. $\endgroup$ – Peter Jun 25 '18 at 13:22
  • 1
    $\begingroup$ This math overflow question answers your question. $\endgroup$ – Sophie Jun 25 '18 at 13:29
  • $\begingroup$ When a is odd then $d=a^2-2b^2$ is odd. $\endgroup$ – sirous Jun 25 '18 at 14:09
0
$\begingroup$

these are numbers that are not divisible by $4$ or by any prime $q \equiv 3,5 \pmod 8.$ You also are throwing out the single factor of $2$ that would otherwise be allowed.

   1 =  1 
   7 = 7
  17 = 17
  23 = 23
  31 = 31
  41 = 41
  47 = 47
  49 = 7^2
  71 = 71
  73 = 73
  79 = 79
  89 = 89
  97 = 97
 103 = 103
 113 = 113
 119 = 7 * 17
 127 = 127
 137 = 137
 151 = 151
 161 = 7 * 23
 167 = 167
 191 = 191
 193 = 193
 199 = 199
 217 = 7 * 31
 223 = 223
 233 = 233
 239 = 239
 241 = 241
 257 = 257
 263 = 263
 271 = 271
 281 = 281
 287 = 7 * 41
 289 = 17^2
 311 = 311
 313 = 313
 329 = 7 * 47
 337 = 337
 343 = 7^3
 353 = 353
 359 = 359
 367 = 367
 383 = 383
 391 = 17 * 23
 401 = 401
 409 = 409
 431 = 431
 433 = 433
 439 = 439
 449 = 449
 457 = 457
 463 = 463
 479 = 479
 487 = 487
 497 = 7 * 71
 503 = 503
 511 = 7 * 73
 521 = 521
 527 = 17 * 31
 529 = 23^2
 553 = 7 * 79
 569 = 569
 577 = 577
 593 = 593
 599 = 599
 601 = 601
 607 = 607
 617 = 617
 623 = 7 * 89
 631 = 631
 641 = 641
 647 = 647
 673 = 673
 679 = 7 * 97
 697 = 17 * 41
 713 = 23 * 31
 719 = 719
 721 = 7 * 103
 727 = 727
 743 = 743
 751 = 751
 761 = 761
 769 = 769
 791 = 7 * 113
 799 = 17 * 47
 809 = 809
 823 = 823
 833 = 7^2 * 17
 839 = 839
 857 = 857
 863 = 863
 881 = 881
 887 = 887
 889 = 7 * 127
 911 = 911
 919 = 919
 929 = 929
 937 = 937
 943 = 23 * 41
 953 = 953
 959 = 7 * 137
 961 = 31^2
 967 = 967
 977 = 977
 983 = 983
 991 = 991
$\endgroup$
  • $\begingroup$ I came to the same result, nevertheless, thank you. Since $d$ must be odd, we can even formulate it simpler : If $d>1$, then all prime factors of $d$ must be of the form $8k\pm 1$. This is also sufficient for the representation, so a pythagorean triple with such a $d$ will actually exist. $\endgroup$ – Peter Jun 25 '18 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.