# Is there a way to find a pythagorean triple so that when you place a given digit before it, it still is a pythagorean triple?

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?

• 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 $221^2=20^2+220^2$. – Ian Miller Sep 21 '16 at 13:53
• 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 $b = 4n + 2$? – Ade Sep 21 '16 at 14:06
• I have a faint intuition that there are no other examples in base 10. Can that be proved? – Greg Martin Sep 21 '16 at 19:02
• @GregMartin Maybe finitely many base $10$ examples with $x,y,z$ not all divisible by $10$? – Robert Israel Sep 22 '16 at 0:37
• @Ade On further searching I did find ones with base 18 as well. – Ian Miller Sep 22 '16 at 2:24

## 2 Answers

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.

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:

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