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In 1918, Norman Alliston noted that the following system of quadratic Diophantine equations \begin{cases} \begin{split} a^2\,\quad+c^2&=u^2\\ b^2\,\quad+c^2&=v^2\\ (a+b)^2+c^2&=w^2 \end{split} \end{cases} has the minimum positive integer solution (a, b, c, u, v, w) = (11, 80, 60, 61, 100, 109).

I haven't found any relevant materials in the library.

Can you tell me how to solve this system in integers?

The links that may be useful are as follows

https://artofproblemsolving.com/community/c3046h1046682

https://artofproblemsolving.com/community/c3046h1054519

\begin{align*} (\,\phantom{0}a\phantom{0},\enspace\phantom{0}b\phantom{0},\enspace\phantom{0}c\phantom{0}\,)&&(\,\phantom{0}u\phantom{0},\enspace\phantom{0}v\phantom{0},\enspace\phantom{0}w\phantom{0}\,)\\ (\,\phantom{0}11,\enspace\phantom{0}80,\enspace\phantom{0}60\,)&&(\,\phantom{0}61,\enspace100,\enspace109\,)\\ (\,\phantom{0}27,\enspace182,\enspace120\,)&&(\,123,\enspace218,\enspace241\,)\\ (\,\phantom{0}38,\enspace319,\enspace360\,)&&(\,362,\enspace481,\enspace507\,)\\ (\,\phantom{0}44,\enspace117,\enspace240\,)&&(\,244,\enspace267,\enspace289\,)\\ (\,\phantom{0}63,\enspace102,\enspace280\,)&&(\,287,\enspace298,\enspace325\,)\\ (\,\phantom{0}90,\enspace119,\enspace120\,)&&(\,150,\enspace169,\enspace241\,)\\ (\,112,\enspace273,\enspace180\,)&&(\,212,\enspace327,\enspace425\,)\\ (\,182,\enspace209,\enspace120\,)&&(\,218,\enspace241,\enspace409\,)\\ \end{align*}

Eufisky
  • 3,237
  • For $a=1,\ldots ,10$ there are only a few primitive solutions to $a^2+c^2=u^2$. In each case it is easy to see that the other equations are contradictory. For example, let $a=3$. Then only $3^2+4^2=5^2$ is possible. For $b^2+4^2=w^2$ again only $b^2=3^2$ and $w^2=5^2$ is possible. But then $(a+b)^2=6^2$ and $c^2=4^2$, whose sum is not a square, contradiction. – Dietrich Burde Apr 10 '20 at 13:55

2 Answers2

1

One solution is:

$a=(x^3-3xy^2)$

$b=(y^3-3x^2y)$

$c=(4xyz)$

Where, $(x^2+y^2=z^2)$

$u=(x^3+5xy^2)$

$v=(5x^2+y^3)$

The above satisisfy's the first two equations. For the third equation take:

$(n^2-m^2)^2+(2mn)^2=(n^2+m^2)^2$

and we get the conditions:

$n^2-m^2=(x+y)(x^2-4xy+y^2)$

$(-2nm)=(4xyz)$

Above two equations are satisfied at, $(x,y,z)=(3,4,5)$ & $(n,m)=(-8,15)$

Hence we have:

$(a,b,c)=(-117,44,240)$ &

$(u,v,w)=(267,244,289)$

Robert
  • 39
1

\begin{cases} \begin{split} a^2\,\quad+c^2&=u^2\\ b^2\,\quad+c^2&=v^2\\ (a+b)^2+c^2&=w^2 \end{split} \end{cases} One of solutions of first and second equations are given below.
$a = p^2-q^2$
$b = p^2q^2-1$
$c = 2pq$
$u = p^2+q^2$
$v = p^2q^2+1$

Substitute above ${a,b,c}$ to third equation, then we get
$$w^2 = (q^4+2q^2+1)p^4+(-2q^4-2)p^2+q^4+2q^2+1$$ If this quartic equation has a rational solution, it is birationally equivalent to the elliptic curve.
For instance, let $q=2/3$ then we get elliptic curve below. $Y^2 = X^3 + X^2 -7924X + 219776$
This curve has rank $2$ with generator $[70, 114],[431, 8778]$.
Thus, we get infinitely many integer solutions.

Example: $q=2/3$. $$[a,b,c], [u,v,w]$$ $$[1485, 595, 468], [1557, 757, 2132]$$ $$[1421, 451, 780], [1621, 901, 2028]$$ $$[1829, 459, 1260], [2221, 1341, 2612]$$ $$[3680, 415, 3432], [5032, 3457, 5343]$$ $$[230585, 97600, 50232], [235993, 109768, 332007]$$ $$[7590044, 3323599, 916320], [7645156, 3447601, 10952043]$$ $$[17359565, 6246355, 7690452], [18986773, 9907573, 24827052]$$ $$[21459680, 8433495, 7347672], [22682728, 11185353, 30782953]$$

Tomita
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