Gap between two Pythagorean triples

Question

Let $s$ and $t$ be two positive integers. I am interested in the following Diophantine system : $$\mathscr{S}(s,t) : \left\{ \begin{array}{l} x^2+y^2=a^2\\ (x-s)^2+(y-t)^2=b^2 \end{array} \right.$$ where $x, y, a$ and $b$ are integers such that $x>s$ and $y>t$.

I would like to know for which couples $(s,t)$ the system $\mathscr{S}(s,t)$ has an infinity of solutions.

Have you encountered this question before? Do you have any references for me?

Thanks in advance.

► Edit : With the help of a computer program, I tested all the pairs $(s,t)$ with $s\leqslant20$ and $t\leqslant20$. In black, the only pairs $(s,t)$ such that $\mathscr{S}(s,t)$ does not seem to have a solution :

Answer 1

I propose a full solution. I claim that all pairs $(s,t)$ allow for infinitely many solutions to the set of equations, except for those blacked out in the table attached to the question.

Suppose without loss of generality that $s \geq t$. I am going to make a case distinction based on the parity of $s$.

FIRST CASE: Suppose first that $s$ is even and write $s = 2k$. I will investigate under which conditions I can find infinitely many solutions with the explicit choice $b = a - t - 2$. Then, the system becomes $$ x^2+y^2 = a^2 \quad \text{and} \quad (x - s)^2 + (y-t)^2 = (a-t-2)^2. $$ This has the somewhat obvious solution $$ (x,y,a,b) = ( 2k, k^2-1, k^2+1, k^2 - t - 1), $$ that does not quite count as $x = s$ in this case. However, I will use it to generate infinitely many more solutions that do count. Expanding the second equation and substituting the first, we obtain $$ x^2+y^2 = a^2 \quad \text{and} \quad 2a(t+2) + s^2 = 2xs + 2yt + 4t + 4. $$ We now use the general solution to the Pythagorean equation to write $$ x = 2uv, \quad y = u^2 - v^2 \quad \text{and} \quad a = u^2+v^2. $$ With this, we have reduced to a single equation reading $$ 2(u^2+v^2)(t+2) + s^2 = 4uvs + 2(u^2-v^2)t + 4t + 4, $$ or equivalently, $$ 4u^2 - 4suv + (4t+4)v^2 = 4t + 4 - s^2. $$ One can rewrite this in the slightly more friendly form $$ [ 2u - sv ]^2 + (4t + 4 - s^2)v^2 = 4t+4 - s^2. $$ Writing $D = s^2 - 4t - 4$, this can be written very succinctly as $$ \text{Nm}( (2u-sv) + v\sqrt{D} ) = -D. $$ Using the solution to our original pair of equations, we can deduce the solution $(u,v) = (k,1)$ to the above equation. If $D$ would be a non-square and $D > 0$, then the group $\mathbb{Z}[\sqrt{D}]$ would have unit rank $1$, so multiplying our solution by any unit of norm $1$ would yield another solution. The infinitude of such units would then generate us our desired infinitude of solutions. One might worry that these solutions will correspond to $x < s$ or $y < t$, but in the limit, this does not happen. Indeed, when $u$ and $v$ grow, we see that $2u - sv$ must tend to $\sqrt{D}v$, which means that $u$ grows about as fast as $v \cdot (s + \sqrt{D})/2 \gg v$, implying that both $x$ and $y$ will grow as $u$ and $v$ do.

It remains to verify under which conditions my two claims are true. Firstly, since $s \geq t$, it holds that $D > 0$ as soon as $s^2 \geq 4s+4$, which certainly holds for $s \geq 6$. Secondly, suppose that $D$ is a square. Again since $s \geq t$, it holds that $D \geq s^2 - 4s - 4 = (s-2)^2 - 8$. Since also $D < s^2$, this only leaves the two possibilities $D = (s-1)^2$ and $D = (s-2)^2$. However, since $D$ is even, only the second case remains, which corresponds to the case that $t = s - 2$.

Suppose that $t = s - 2$ and apply the argument above with $t$ and $s$ swapped. As before, we are then done as soon as $D = (s-2)^2 - 4s - 4 = s^2 - 8s$ is positive and a non-square. It is positive as soon as $s \geq 10$. It is then also a non-square, as $D = (s - 4)^2 - 16$ and the only even squares with difference $16$ are $16$ itself and $0$.

Combining all the arguments above, we conclude that, if $s \geq t$ and $s$ is even, we can find infinitely many solutions as soon as $s \geq 10$, and if $s \in \{ 6, 8 \}$, we can find then for any $t \neq s - 2$.

SECOND CASE: Suppose now that $s \geq t$ and that $s$ is odd, so write $s = 2k+1$. This time, I investigate when I can find infinitely many solutions with the explicit choice $b = a - t - 1$. Indeed, I claim that the pair of equations $$ x^2+y^2 = a^2 \quad \text{and} \quad (x - s)^2 + (y-t)^2 = (a-t-1)^2 $$ admits the again somewhat silly solution $$ (x,y,a,b) = ( 2k+1, 2k(k+1), 2k^2+2k+1, 2k^2 + 2k - t). $$ If we now use the substitutions the other way around, $$ x = u^2-v^2, \quad y = 2uv \quad \text{and} \quad a = u^2+v^2, $$ we can completely similarly deduce the equation $$ [ (t+1-s)u - tv ]^2 + (2t + 1 - s^2)v^2 = (2t+1-s^2)(t+1-s)/2. $$ to which the non-trivial solution $(u,v) = (k+1, k)$ exists. Again, this is a norm equation in the ring $\mathbb{Z}[\sqrt{D}]$, provided that $D = s^2 - 2t - 1$ is positive and a non-square. We again find infinitely many solutions and to ensure that $x$ and $y$ keep growing too, we must check the growth rate of $u$ with respect to $v$. If $t + 1 - s < 0$, we check that $u$ will grow at the same pace as $v \cdot (\sqrt{D} - t) / (s - t - 1) \gg v$, ensuring the correct growth as soon as $\sqrt{D} > s - 1$, or equivalently, $ t < s - 1 $, which is exactly what we assumed. It remains to check the cases that $s = t$ or $s = t + 1$. In the former case, we see that $u$ grows as $v \cdot (\sqrt{D}+t) \gg v$; we postpone the latter case momentarily, as it turns out we will use a different argument there.

Let us investigate when this happens. Firstly, since $s \geq t$, it holds that $D > 0$ as soon as $s^2 > 2s+1$, which happens as soon as $s \geq 3$. Secondly, suppose $D$ is a square. Again, since $s \geq t$, it holds that $D \geq s^2 - 2s - 1$. Since also $D < s^2$, this forces $D = (s-1)^2$ and as such, $t = s - 1$.

Suppose that we are in the case $t = s - 1$. Then we can use the strategy for even $t$ above, to conclude that we can still find infinitely many solutions as soon as $(s-1)^2 > 4s + 4$ and $D = (s-1)^2 - 4s - 4$ is a non-square. The first condition is satisfied as soon as $s \geq 7$ and for the second, we note that $D = (s - 3)^2 - 12$, and the only squares of difference $12$ are $16$ and $4$, which corresponds to $s = 7$.

Combining the above, we conclude that if $s \geq t$ and $s$ is odd, we can find infinitely many solutions as soon as $s \geq 7$, and if $s \in \{ 3, 5, 7 \}$, then we can do it too unless $t = s - 1$.

CONCLUSION: If one now carefully reads the two conclusions reached above, it follows that the only pairs $(s,t)$ with $s \geq t$ for which we could not possibly find infinitely many solutions are: $$ (1,1), (2,1), (2,2), (3,2), (4,1), (4,2), (4,3), (4,4), (5,4), (6,4), (7,6),(8,6). $$ Of these, it turns out that $(4,1)$, $(4,3)$ and $(8,6)$ do allow infinitely many solutions. Indeed, for the first, the discriminant $D = s^2 - 4t - 4$ is actually still positive. For the latter two, we employ the observation that $4^2 + 3^2 = 5^2$ and $8^2+6^2 = 10^2$. These generate the infinite families of solutions with $(x,y) = (4c, 3c)$ and $(x,y) = (8c,6c)$ respectively.

Finally, I claim that the remaining pairs, $$ (1,1), (2,1), (2,2), (3,2), (4,2), (4,4), (5,4), (6,4), (7,6), $$ really do not admit infinitely many solutions. The argument contains too many case distinctions for me to write it all out here, but let me give you the tools to verify this for yourself. Writing $b = a - n$, my arguments above generate bijections between solutions $(x,y,a,b)$ to the pair of equations in question and solutions $(u,v)$ to the norm equation $$ [ (n-s)u - tv ]^2 + (n^2 - s^2 - t^2)v^2 = (n^2 - s^2 - t^2)(n-s)/2, $$ or with $s$ and $t$ swapped when parametrising the Pythagorean triples differently. It is important to note here that in order to obtain infinitely many solutions from this equation, the quadratic form on the left hand side must be indefinite. Hence we need only consider $0 < n < \sqrt{s^2 + t^2}$. The right hand side must also be integral. These leave only finitely many choices for $n$. Explicitly, one deals with these cases as follows:

For $(1,1), (2,1), (2,2), (4,2)$, there aren't even any valid $n$ to consider because the inequalities are too strict to obtain an integral right hand side for any $n$, so we are immediately done. For $(3,2), (4,4), (7,6)$, there are various $n$ yielding a sensible equation, but none of these turn out to have a solution. Finally, most interestingly, there are the cases of $(5,4)$ and $(6,4)$. In these cases, there are actually infinitely many pairs of solutions $(u,v)$ for some values of $n$. However, if one would work out the growth rates of $u$, they all turn out to be $\ll 0.85v$ and as such, all these infinite solutions will yield negative $x$ or $y$, hence not contributing actual solutions to the problem at hand.

Answer 2

Comment: $t$ can be fixed but if $s$ is a number in an arithmetic progression then the system can have infinitely many solutions.For example:

$a=2i+1=3, 5, 7, 9 . . .$, where $t=2$

$b=2i(i+1)= 4, 12, 24, 40, . . .$ where $s=8, 12, 16, 20, 24, . . .$

$c=2i(i+1)+1=5, 13, 25, 41, . . .$

Another example:

$a=4i=4, 8, 12, 16, . . .$ where $t=4$

$b=4i^2-1=3, 15, 35, 63, . . .$ where $s=12, 20, 28, 36, . . .$

$c=4i^2+1=5, 17, 37, 65, . . .$

Update: We use quadratic residue; suppose we have"

$x^2+y^2=a^2$

$x^2\equiv r_x \mod m=k m +r_x\Rightarrow (x+ s d)^2\equiv r_x\bmod m$

Let $d=u-1$ we may write:

$x+sd= x+su-s\rightarrow (x+su-s)^2\equiv r_x\bmod m$

we assume $x+su=X$ so we get:

$(X-s)^2\equiv r_x\bmod m=m k_1+r_x$

Similarly for y we may write:

$y^2\equiv r_y \mod n\Rightarrow (x+ t d_1)^2\equiv r_y\bmod n$

Let $d_1=v-1$ we may write:

$y+td_1= y+tv-t\rightarrow (y+tv-t)^2\equiv r_y\bmod n$

we assume $y+tv=Y$ so we get:

$(Y-t)^2\equiv r_y\bmod n=n k_2+r_y$

That is for fixed $s$ and $t$ we can have new x (X), and new y(Y) such that:

$(X-s)^2+(Y-t)^2=b_1^2$

provided that:

$mk_1+nk_2+r_x+r_y=b_1^2$

That is for fixed $s$ and $t$ we may have infinitely many solutions for x, y, and b.

Example: Take $5^2+12^2=13^2$

$(5+3p)^2=\equiv 2\bmod 3=3k+2$

$(12+5q)^2=2\bmod 5=5k_1+2$

Take $b^2=25$ , we can write:

$3k+5k_1+4=25\Rightarrow 3k+5k_1=21$

Which has infinite solutions in $\mathbb Z$. Take one solution, for example $k=2$, $k_1=3$, we have:

$(X-3)^2+(Y-5)^2=25=5^2$

$X-3=3\rightarrow X=6$; $Y-5=4\rightarrow Y=9$

such that:

$(6-3)^2+(9-5)^2=5^2$

Another example :

Take $b^2=61^2$ we have:

$3k+5k_1+4=61^2=3721\Rightarrow 3k+5k_1=3717$

Take one solution $k=239$, $k_1=600$ such that:

$3\times 239+ 5\times 600+4=3721=61^2$

Compare with:

$11^2+60^2=61^2$

and get $X=14$ and $Y=65$ such that:

$(14-3)^2+(65-5)^2=61^2$

Answer 3

We can see a pattern for $\,(s,t)\,$ in the sets below. The value of $\,s\,$ is fixed in any given row and the value of $\,t\,$ is fixed in any given column but both are never fixed at the same time.

\begin{align*} &A=(2n-1)^2+2(2n-1)k\\ &B=\phantom{(2n-1)^2+{}} 2(2n-1)k+2k^2\\ &C=(2n-1)^2+2(2n-1)k+2k^2 \end{align*}

$$\begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 &k=3 & k=4 & k=5\\ \hline Set_1&3,4,5 &5,12,13&7,24,25&9,40,41&11,60,6\\ \hline Set_2&15,8,17&21,20,29 &27,36,45 &33,56,65&39,80,89\\ \hline Set_3&35,12,37&45,28,5&55,48,73&65,72,97&75,100,125\\ \hline Set_{4}&63,16,65&77,36,85&91,60,109&105,88,137&119,120,169\\ \hline Set_{5}&99,20,101&117,44,125&135,72,153&153,104,185&171,140,221\\ \hline \end{array}$$

However, if we consider only the first set, i.e. $\,n=1,\,$ then $$A = 2 k + 1\qquad B = 2 k^2 + 2 k\qquad B = 2 k^2 + 2 k + 1$$ For any $Set_1$ triple, $\, A = 2 k + 1\implies k=\dfrac{A-1}{2}$ and reducing $\,k\,$ by $1$ yields $$A=2k-1\quad B=2 k^2 - 2 k\quad C=2 k^2 - 2 k + 1$$

So, for example, the triple \begin{align*} T_n&=(13,84,85)\\\implies k&=\dfrac{13-1}{2}=6\\\implies T_{n-1}&=(12-1,72-12,72-12+1\\ &=(11,60,61) \end{align*} and the algorithm works for any of the infinite members of $Set_1$ except $(3,4,5)$

A similar algorithm can be found for any other given set.

Answer 4

Too big for a comment:

Ony one of (s,t) can be fixed at a time.

Per a previous OP comment question, there cannot be a function $\space\mathscr{S}(18,19)\space $ because $\,t\,$ must be a multiple of $\,4.\quad$ However, $\space\mathscr{S}(18,4p)\space $ is possible (where $p\in\mathbb{N}, p\ge5$) using $Set_5$ of the table below. $$\begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 &k=3 & k=4 & k=5\\ \hline Set_1&3,4,5 &5,12,13&7,24,25&9,40,41&11,60,6\\ \hline Set_2&15,8,17&21,20,29 &27,36,45 &33,56,65&39,80,89\\ \hline Set_3&35,12,37&45,28,5&55,48,73&65,72,97&75,100,125\\ \hline Set_{4}&63,16,65&77,36,85&91,60,109&105,88,137&119,120,169\\ \hline Set_{5}&99,20,101&117,44,125&135,72,153&153,104,185&171,140,221\\ \hline \end{array}$$

To find a triple $\,T_{n-1}\,$ from any $\,T_n\,$ in this set except $(99,20.101)$, find the $\,k$-value of $\,T_{n-1}\,$ using $\,k = \dfrac{A_{n} - 81}{18}$ and generate the new triple using the following formula.

$$ A=9 (2 k + 7)\qquad B=2 (k^2 + 7 k - 8)\qquad C=2 k^2 + 14 k + 65$$

For example, \begin{align*}(153,104,185)\implies \,k_{n-1} = \dfrac{153 - 81}{18}=4\\ \end{align*}

\begin{align*} A_{n-1}=9 (2 (4) + 7)=135\\ B_{n-1}=2 ((4)^2 + 7 (4) - 8)=72\\ C_{n-1}=2 (4)^2 + 14 (4) + 65=153 \end{align*}

Answer 5

Let us consider the possible solutions of the given system in the form of

$$\begin{cases} x^2+y^2=a^2\\ (x-s)^2+(y-t)^2=b^2\\ x+iy=p^2-q^2+2ipq =(p+iq)^2\\ a=p^2+q^2=|x+iy|\\ \sigma+i\tau =(x-s)+i(y-t)=u^2-v^2+2iuv=(u+iv)^2\\ b=|\sigma+i\tau|=u^2+v^2=|u+iv|^2\ge5, \end{cases}\tag1$$ which corresponds with the case of the even values of $\,s\,$ and $\,t\,.$

Taking in account $\,(1),\,$ one can get $$s+it=x+iy -(\sigma+i\tau)=(p+iq)^2-(u+iv)^2= \big(p+u+i(q+v)\big)\big(p-u+i(q-v)\big),$$ $$s^2+t^2=|s+it|^2=\big((p+u)^2+(q+v)^2\big)\big((p-u)^2+(q-v)^2\big)$$ $$=(a+b+2pu+2qv)(a+b-2pu-2qv)=(a+b)^2-4(pu+qv)^2,$$

$$\begin{cases} s^2+t^2=(a+b)^2-4(pu+qv)^2\\[4pt] (p^2+q^2)^2=a^2=x^2+y^2\\[4pt] (u^2+v^2)^2=b^2=x^2-2sx+s^2+y^2-2yt+t^2\\[4pt] b^2=\big(a^2-2\left(sx+t\sqrt{a^2-x^2}\,\right)s^2+t^2\big). \end{cases}\tag2$$ Applying $(2),$ one can get: $$\left(sx+t\sqrt{a^2-x^2}\,\right)'=s-\dfrac{tx}{\sqrt{s^2+t^2}}=0,\quad x=\dfrac{as}{\sqrt{s^2+t^2}},$$ $$b^2\ge \big(a^2-2\left(sx+t\sqrt{2a^2-x^2}\,\right)s^2+t^2\big) \bigg|_{\large x =\frac{a s}{\sqrt{s^2+t^2}}}$$ $$=a^2-2a\sqrt{s^2+t^2}+s^2+t^2 = (a-\sqrt{s^2+t^2})^2.$$ Since $$a-\sqrt{s^2+t^2} = \sqrt{x^2+y^2}-\sqrt{s^2+t^2\,}\ge0,$$ then \begin{cases} b\ge a-\sqrt{s^2+t^2}\\[4pt] s^2+t^2=(a+b)^2-4(pu+qv)^2\\[4pt] 4(pu+qv)^2\ge(a+b)^2-(a-b)^2= 4ab = 4(p^2+q^2)(u^2+v^2). \end{cases} At the same time, from the CS inequality follows that $$4(pu+qv)^2\le4(p^2+q^2)(u^2+v^2),$$ i.e. there is a strict equality, with the necessary condition $$\dfrac pu = \dfrac qv = \dfrac mn,$$ $$p=m\alpha,\; q= n\alpha,$\quad u=m\beta,\;v=n\beta.$$

Therefore, $$\begin{cases} x+iy=(m^2-n^2+2imn)\,\alpha^2\\ a=(m^2+n^2)\alpha^2,\\ \sigma+i\tau= (m^2-n^2+2imn)\beta^2\\ b=(m^2+n^2)\beta^2\\ s+it=(m^2-n^2+2imn)(\alpha^2-\beta^2)\\ s^2+t^2=(m^2+n^2)^2(\alpha^2-\beta^2)^2 \end{cases}$$ which defines all possible solutions of the system $(1).$ wherein $$\begin{cases} x^2+y^2=a^2\\ (x-s)^2+(y-t)^2=b^2, \end{cases}$$ However, our task is to point the fixed pair $\;(s,t),\;$ which provides infinity many solutions.

Chosen $\;\alpha\,$ and $\,\beta\,$ as hypotenuse and leg of the Pythagorean triangle, easily to eliminate the second factor in the expression for $\;s+it.\,$ At the same time, there are not possibilities to fix the first factor.

The similar fatal problem should not be fixed, when both $\,s,t\;$ are odd. So, the final answer is NO.