9
$\begingroup$

For example, in base 10:

$$5^2 + 12^2 = 13^2$$

And when I put a one before each number, the equality still holds:

$$15^2 + 112^2 = 113^2$$

So my question is, in a given base, is there a way to get pythagorean triples that still hold when a given digit precedes each?

Bonus question: And if there is, are there infinitely many?

$\endgroup$
  • 1
    $\begingroup$ I did a search of bases 2 through 15 looking at Pythagorean triples $(a^2+b^2,a^2-b^2,2ab)$ up to $a=1000$ and they were surprisingly rare. I got $1b^2_{14}=7^2_{14}+1a^2_{14}$ paired with $11b^2_{14}=17^2_{14}+11a^2_{14}$ which in base ten is $25^2=7^2+24^2$ paired with $221^2=20^2+220^2$. $\endgroup$ – Ian Miller Sep 21 '16 at 13:53
  • $\begingroup$ Thanks!, I see that your answer is the case n = 3 of the answer down below, did you find any other triples with a relation different than $b = 4n + 2$? $\endgroup$ – Ade Sep 21 '16 at 14:06
  • $\begingroup$ I have a faint intuition that there are no other examples in base 10. Can that be proved? $\endgroup$ – Greg Martin Sep 21 '16 at 19:02
  • 1
    $\begingroup$ @GregMartin Maybe finitely many base $10$ examples with $x,y,z$ not all divisible by $10$? $\endgroup$ – Robert Israel Sep 22 '16 at 0:37
  • 1
    $\begingroup$ @Ade On further searching I did find ones with base 18 as well. $\endgroup$ – Ian Miller Sep 22 '16 at 2:24
8
$\begingroup$

Here's one infinite family of solutions. Let the base be $b=4n+2$, and take the Pythagorean triple $$ x = 2n+1,\ y = 2n^2 + 2n,\ z = 2 n^2 + 2 n + 1 $$ Note that $1 \le x < b$, $b \le y, z < b^2$, so $x$ has one digit in base $b$ while $y$ and $z$ have two. Putting a $1$ before each in base $b$, we get $$ x' = b + x = 6n+3,\ y' = b^2 + y = 18 n^2 + 18 n + 4,\ z' = b^2 + z = 18 n^2 + 18 n + 5$$ which is again a Pythagorean triple. The case $n=2$ is your example above.

EDIT: Somewhat more generally, start off with a Pythagorean triple of the form $$(x,y,z) =(m^2-n^2,\; 2mn,\; m^2+n^2)$$ with $m>n$, and suppose in base $b$, $x$ has one digit while $y$ and $z$ each have two.

Prepending the digit $1$ gives $$(x',y',z') = (b+m^2-n^2,\; b^2+2mn, \; b^2+m^2+n^2)$$ For this to be a Pythagorean triple, you need $$ 0 = x'^2 + y'^2 - z'^2 = b^2(1-2(m-n)^2)+2b(m^2-n^2)$$ From this we see that $2(m-n)$ must divide $b$, so let's take $b= 2 (m-n)r$.

After factoring out $4 (m-n)^2 r$ we then get the equation $$ 2(m-n)^2r - m - n - r = 0$$ Let $m+n = s$ and $m-n = t$, so $m = (s+t)/2$, $n = (s-t)/2$. The equation then becomes $2rt^2 - r - s = 0$. Thus $s = 2 rt^2 - r$. Our solution is now $$ x = 2rt^3-rt,\; y=2 r^2 t^4 - 2 r^2 t^2 + \dfrac{r^2-t^2}{2},\; z = 2 r^2 t^4 - 2 r^2 t^2 + \dfrac{r^2+t^2}{2},\; b = 2 r t$$ But we wanted $x < b$, so $2rt^3 - rt < 2rt$, i.e. $2 t^2 < 3$. Thus we need $t=1$. And $r$ must be odd for $y$ and $z$ to be integers. Taking $r=2k+1$ leaves us with my solution (with $k$ instead of $n$).

EDIT: There are other solutions. Some with base not of the form $4n+2$ are $$ \matrix{b = 12 & x = 442 & y = 120 & z = 458 & x' = 2170 & y' = 264 & z' = 2186\cr b = 12 & x = 210 & y = 72 & z = 222 & x' = 1938 & y' = 216 & z' = 1950\cr b = 12 & x = 120 & y = 160 & z = 200 & x' = 840 & y' = 8800 & z' = 8840\cr b = 16 & x = 494 & y = 192 & z = 530 & x' = 12782 & y' = 960 & z' = 12818\cr b = 16 & x = 504 & y = 128 & z = 520 & x' = 4600 & y' = 384 & z' = 4616\cr b = 20 & x = 546 & y = 360 & z = 654 & x' = 72546 & y' = 3960 & z' = 72654\cr b = 20 & x = 1326 & y = 360 & z = 1374 & x' = 25326 & y' = 1560 & z' = 25374\cr b = 20 & x = 884 & y = 240 & z = 916 & x' = 16884 & y' = 1040 & z' = 16916\cr b = 20 & x = 442 & y = 120 & z = 458 & x' = 8442 & y' = 520 & z' = 8458\cr b = 20 & x = 10 & y = 24 & z = 26 & x' = 70 & y' = 1224 & z' = 1226\cr b = 20 & x = 9999 & y = 200 & z = 10001 & x' = 489999 & y' = 1400 & z' = 490001\cr b = 20 & x = 990 & y = 200 & z = 1010 & x' = 8990 & y' = 600 & z' = 9010\cr b = 21 & x = 561 & y = 252 & z = 615 & x' = 37605 & y' = 2016 & z' = 37659\cr b = 24 & x = 836 & y = 480 & z = 964 & x' = 125252 & y' = 5664 & z' = 125380\cr b = 24 & x = 35 & y = 12 & z = 37 & x' = 1763 & y' = 84 & z' = 1765\cr b = 28 & x = 14 & y = 48 & z = 50 & x' = 98 & y' = 2400 & z' = 2402\cr b = 32 & x = 63 & y = 16 & z = 65 & x' = 3135 & y' = 112 & z' = 3137\cr b = 36 & x = 18 & y = 80 & z = 82 & x' = 126 & y' = 3968 & z' = 3970\cr b = 40 & x = 20 & y = 48 & z = 52 & x' = 300 & y' = 11248 & z' = 11252\cr b = 40 & x = 99 & y = 20 & z = 101 & x' = 4899 & y' = 140 & z' = 4901\cr }$$ There are also other solutions in base $10$. Besides those obtained by multiplying $x,y,z$ by the same power of $10$, we have $$ \matrix{x = 1045 & y=600 & z=1205 & x' = 21045 & y' = 2600 & z' = 21205\cr x = 11242 & y=600 & z=11258 & x' = 211242 & y' = 2600 & z' = 211258\cr x = 12495 & y=500 & z=12505 & x' = 112495 & y'=1500 & z'=112505\cr x = 15675 & y=9000 & z=18075 & x' = 315675 & y'=39000 & z'=318075\cr x = 16863 & y=900 & z=16887 & x' = 316863 & y'=3900 & z'=316887\cr }$$ I don't know if these are in infinite families.

$\endgroup$
2
$\begingroup$

Inspired by Robert's analysis I decided to find a different family.

This works for bases of the form $b=4mn$ with $n>4$ and $m>0$.

Take the triple $(2mn,m(n^2-1),m(n^2+1))$

Clearly $2mn<4mn$ so its one digit long. Also $4mn<m(n^2-1)<m(n^2+1)<(16m^2n^2)$ so those two are two digits long.

If you put the digit $4m-1$ (which is a digit as $4m-1<4mn$) in front of each number you get the triple:

$$\bigg((4m-1)\cdot(4mn+2mn),(4m-1)\cdot(4mn)^2+m(n^2-1),(4m-1)\cdot(4mn)^2+m(n^2+1)\bigg)$$ $$=\bigg(16m^2n-2mn,64m^3n^2-16m^2n^2+mn^2-m,64m^3n^2-16mn^2+mn^2+m\bigg)$$

Which satisifies the Pythagorean Identity.

With $m=1$ this leads to Robert's solutions of:

$$\matrix{b = 20 & x = 10 & y = 24 & z = 26 & x' = 70 & y' = 1224 & z' = 1226\cr b = 24 & x = 35 & y = 12 & z = 37 & x' = 1763 & y' = 84 & z' = 1765\cr b = 28 & x = 14 & y = 48 & z = 50 & x' = 98 & y' = 2400 & z' = 2402\cr b = 32 & x = 63 & y = 16 & z = 65 & x' = 3135 & y' = 112 & z' = 3137\cr b = 36 & x = 18 & y = 80 & z = 82 & x' = 126 & y' = 3968 & z' = 3970\cr b = 40 & x = 99 & y = 20 & z = 101 & x' = 4899 & y' = 140 & z' = 4901\cr}$$

And with $m=2$ this leads to solutions (Robert had first one) of:

$$\matrix{b = 40 & x = 20 & y = 48 & z = 52 & x' = 300 & y' = 11248 & z' = 11252\cr b = 48 & x = 24 & y = 70 & z = 74 & x' = 360 & y' = 16198 & z' = 16202\cr b = 56 & x = 28 & y = 96 & z = 100 & x' = 420 & y' = 22048 & z' = 22052\cr b = 64 & x = 32 & y = 126 & z = 130 & x' = 480 & y' = 28798 & z' = 28802\cr b = 72 & x = 36 & y = 160 & z = 164 & x' = 540 & y' = 36448 & z' = 36452\cr b = 80 & x = 40 & y = 198 & z = 202 & x' = 600 & y' = 44998 & z' = 45002\cr}$$

$\endgroup$

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.