# Finding $n$ satisfying that there is no set $(a,b,c,d)$ such that $a^2+b^2=c^2$ and $a^2+nb^2=d^2$

Let us consider $n\ge 3\in\mathbb N$ which satisfy the following condition.

Condition : There exist no set of four non-zero integers $(a,b,c,d)$ such that $$a^2+b^2=c^2\ \ \text{and}\ \ a^2+nb^2=d^2.$$

Then, here is my question.

Question : How can we find every such $n$?

Remark : Please note $n\ge 3$. This is because it is known that $n=2$ satisfies the condition. ($a^2+b^2=c^2,a^2+2b^2=d^2\Rightarrow c^2-b^2=a^2,c^2+b^2=d^2\Rightarrow c^4-b^4=(ad)^2$ and see, for example, here)

The followings are the examples of $n$ which do not satisfy the condition.

• For $n=4k^2+5k+1\ (n=10,27,52,85,126,\cdots)$, take $(a,b,c,d)=(3,4,5,8k+5).$

• For $n=4k^2+3k\ (n=7,22,45,76,115,162,\cdots)$, take $(a,b,c,d)=(3,4,5,8k+3).$

• For $n=9k^2+10k+1\ (n=20,57,112,185,\cdots)$, take $(a,b,c,d)=(4,3,5,9k+5).$

• For $n=9k^2+8k\ (n=17,52,105,176,265,\cdots)$, take $(a,b,c,d)=(4,3,5,9k+4).$

We know that $(a,b,c)$ is a Pythagorean triple and we can see that $$\text{(a,b,d)=\left((s^2-nt^2)u,2stu,(s^2+nt^2)u\right)\  satisfy \ a^2+nb^2=d^2}.$$ However, I don't have any good idea to find such $n$. Can anyone help?

Added : A user individ found that if there are integers $p,s,t$ such that $$(a,b,c,d,n)=(p-s,2t,p+s,\mp 2n+p+s\pm 2,(p\pm 1)(s\pm 1))\ \ \text{and}\ \ ps=t^2,$$ then the $n$ does not satisfy the condition.

However, this does not say anything about $n$ which satisfy the condition. We still don't know if each of $n=3,4,5,6,8$, for example, satisfies the condition.

• the answer that was accepted at the previous question is incorrect. $1 + 49 = 2 \cdot 25$ Commented Feb 14, 2015 at 19:31
• @WillJagy: Thank you for pointing it out. I changed the link to another one. Commented Feb 14, 2015 at 19:39
• A quick find counter-example is: $$f(4,1):15^2+8^2=17^2\quad\text{and}\quad f(8,7):15^2+112=113^2\qquad \text{Note:}112=14*8$$ Commented Jun 23, 2019 at 2:49

The Diophantine system $a^2 + b^2 = c^2$, $a^2 + nb^2 = d^2$ gives rise to the elliptic curve $$E_n : y^2 = x (x+1) (x+n).$$ As is often the case for families of elliptic curves with a simple equation, this family has a long history, and still resists a complete answer even though we've made considerable progress since the 19th-century (and earlier) work reported in Dickson's History of the Theory of Numbers, Vol.2: Diophantine Analysis (see the section on "concordant forms"). It takes Cremona's program mwrank less than a minute to find that the sequence of integers $n > 1$ for which there are no solutions in nonzero $a,b,c,d$ begins

2, 3, 4, 5, 6, 8, 9, 12, 13, 14, 15, 16, 18, 19, 21, 25, 26, 28, 29, 32, 33, 35, 36, 37, 38, 39, 40, 43, 44, 46, 48, 51, 54, 55, 56, 62, 63, 64, 65, 66, 67, 69, 70, 73, 75, 78, 80, 81, 84, 87, 88, 89, 91, 95, 96, 98, … .

These are also the integers $n \in [2,100]$ for which $E_n$ has rank zero (the group of torsion points is isomorphic with $({\bf Z}/2{\bf Z}) \times ({\bf Z}/4{\bf Z})$ if $n$ is a square, and $({\bf Z}/2{\bf Z}) \times ({\bf Z}/2{\bf Z})$ otherwise). In each case this can be proved with a Fermat-style "$2$-descent". For each integer $n \in [2,100]$ not in that list, $E_n$ has rank $1$, except for $n=31$, $52$, $71$, $74$, $79$ for which $E_n$ has rank $2$. [The OEIS seems to contain neither the above sequence of $n$, nor the complementary sequence for which there are nonero solutions, nor the same sequences with $n$ replaced by $n-1$ which corresponds to $y^2 = x (x-1) (x+n)$.] For $n \leq 100$ the minimal solution gets as large as $$(a,b,c,d) = (2873161, 2401080, 3744361, 22062761)$$ for $n=83$. Extending the computation to $n \leq 200$ finds the minimal solution $$(a,b,c,d) = (13265620549, 6755532420, 14886702349, 80460628949),$$ for $n=138$, and also two cases ($n=124$ and $n=195$) where $E_n$ seems to have rank zero but mwrank cannot prove it.

To bring the system $a^2 + b^2 = c^2$, $a^2 + nb^2 = d^2$ into the "Weierstrass form" $y^2 = x (x+1) (x+n)$, start with the parametrization $(a:b:c) = (t^2-1 : 2t : t^2+1)$ of $a^2+b^2=c^2$ up to scaling, and find $a^2+nb^2 = t^4 + 2(2n-1) t^2 + 1$; if this is to be a square, we can write it as $(t^2-(2x+1))^2$ for some $x$ to obtain $(n+x) t^2 = x (x+1)$; given $x$ this has a solution $t$ iff $x (x+1) (x+n)$ is a square. The solutions with $b=0$ come from the subgroup $E_n[2]$ consisting of the "point at infinity" and the three $2$-torsion points at which $y=0$; if $n$ is a square, say $n=m^2$, then there are also solutions with $a=0$, and these come from $4$-torsion points such as $(x,y) = (m,m^2+m)$. From a non-torsion point $(x,y)$ we can recover $(a:b:c:d)$ by reversing the above procedure, or by doubling $(x,y)$ in the group law: the new point will have $x$-coordinate $(a/b)^2$, with each of the factors $x$, $x+1$, $x+n$ of $y^2$ being square separately.

• I'm very glad to read this answer! Thank you so much! Two questions. Is the number of $n$ satisfying the condition infinite? And what is the known biggest such $n$? Commented Feb 21, 2015 at 16:36
• You're welcome. The usual expectation for such problems is that about half the values of $n$satisfy the condition but it's very hard to prove even that there are infinitely many. Here it seems that if $p \equiv 3 \bmod 4$ and $p,2p-1$ are both primes then $n=2p$ yields a curve $E_n$ of rank zero and thus satisfies your condition. This should already yield infinitely many examples; the proof is again expected to be a very hard problem (similar to the infinitude of Sophie Germain primes), but at least it's easy to find lots of large numerical examples such as $n = 10000000502$. Commented Feb 21, 2015 at 20:28
• @NoamD.Elkies: Incidentally, the solvable case $n=52$ has a special use for equal sums of 10th powers. See my answer below. :) Commented May 6, 2015 at 15:36

For the system of equations:

\left\{\begin{aligned}&a^2+b^2=c^2\\&a^2+qb^2=w^2\end{aligned}\right.

If you can decompose the coefficient multipliers as follows: $q=(p\pm1)(s\pm1)$

Their work squares: $ps=t^2$

Then decisions can be recorded. $$a=p-s$$ $$b=2t$$ $$c=p+s$$ $$w=\mp2q+p+s\pm2$$

You can add another simple option. If the ratio can be written as: $$q=2t^2-1$$

Then decisions can be recorded.

$$a=t^2-1$$

$$b=2t$$

$$c=t^2+1$$

$$w=3t^2-1$$

• I'm glad to read this answer. This does show a condition of $n$ which do not satisfy the condition. By the way, double-sign corresponds, right? Then, the case $q=(p+1)(s+1)$ does not happen because of $a,b,c,w\gt 0$ in this question. Anyway, thank you so much. Commented Feb 15, 2015 at 10:28
• @mathlove I did not understand. What's the difference what the sign at the $a,b,c,w$ ? This approach allows us to consider other solutions. Commented Feb 15, 2015 at 10:33
• Do you mean the condition $a,b,c,d\gt 0$ is strange? Commented Feb 15, 2015 at 10:46
• @mathlove it is necessary to consider all the options. As more and less than zero. Commented Feb 15, 2015 at 10:48
• I think I misunderstood something. I agree that your approach allows us to consider other solutions. Thank you. Commented Feb 15, 2015 at 11:08

What you are looking for are integers that are not concordant forms. To quote from the link, a concordant form is an integer triple $a,b,n$ such that,

$$a^2+b^2 = c^2$$ $$a^2+nb^2 = d^2$$

and integer $c,d$. As early as 1857, the list of solvable $n<100$ was given (though missing three terms, in blue):

$$\small n=1, 7, 10, 11, 17, 20, 22, 23, 24, 27, 30, 31, 34, 41, 42, 45, \color{blue}{47}, 49, 50, 52, \color{blue}{53}, 57, 58, 59, 60, 61, 68, 71, 72, 74, 76, 77, 79, 82, \color{blue}{83}, 85, 86, 90, 92, 93, 94, 97, 99, 100$$

For some bizarre reason, this is not yet in the OEIS. (Anyone care to submit it?)

Your question about the $n$ of non-concordant forms would simply be its complement, and given by Elkies' answer as $\small n = 2, 3, 4, 5, 6, 8, 9, 12, 13, 14, 15,\dots$

P.S. Incidentally, the concordant form with $n=52$ has a special use in equal sums of like powers. Let,

$$a^2+b^2 = c^2$$ $$a^2+52b^2 = d^2$$

with initial solution $a,b,c,d = 3,4,5,29$, and an infinite more. Then,

$$(8b)^k + (5a-4b)^k + (-a-2d)^k + (a-2d)^k + (-5a-4b)^k + (-12b+4c)^k + (12b+4c)^k =\\ (4a+8b)^k + (3a-2d)^k + (-3a-2d)^k + (-4a+8b)^k + (-16b)^k + (a+4c)^k + (-a+4c)^k$$

for $k=1,2,4,6,8,10,$ found by J. Wroblewski and yours truly.

• I think the mathworld link misunderstands the classical terminology: two quadratic forms $Q(a,b)$, $Q'(a,b)$ are concordant if they can be made squares simultaneously with $(a,b) \neq (0,0)$. It's true that most of the classical work is on the special case $Q = a^2 + b^2$, $Q' = a^2 + nb^2$; but the relevant chapter in Dickson II also gives the example of the forms $a^2+2b^2$ and $a^2+3b^2$ which are not concordant. Commented May 6, 2015 at 15:55
• @NoamD.Elkies: I see. What I also find surprising is, considering how extensive and eclectic the OEIS is, it doesn't have the sequence for either the solvable and non-solvable $n$. And anyone submitting it has to contend with that terminology issue you pointed out. Commented May 6, 2015 at 16:08