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I'm using the quadratic Diophantine equations to solve for two integer triangles of longest side $25$. It's been shown that for $a^2+b^2=c^2$, which goes to $x^2+y^2=1$ where $x=\frac ac, y=\frac bc, a=t^2−1, b=2t, c=t^2+1$.

If I want to solve for $c=25$, how will I go about doing this?

Just letting $c=25$ gives me an irrational answer for $b$ after having solved for $t$. I'm a little confused.

Thanks for your help.

Mark Bennet
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user2768428
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  • Isn't the formula $m^2 - n^2, 2mn, m^2 + n^2$? Also this formula only works for primitive triples – MT_ Apr 14 '14 at 13:25
  • Could you please explain how you arrived at that, and then how to obtain integer solutions? (ie 7,24,25 and 15,20,25) Thanks very much. The formula I have is primitive I know – user2768428 Apr 14 '14 at 13:29
  • I'm saying your formula isn't correct. Sure, if you have $a, b, c$ in that form it will be a pythagorean triple, but the strength of the other formula I stated is its biconditionality. Where did you learn/find the formula you have? – MT_ Apr 14 '14 at 15:15
  • Similar question: http://math.stackexchange.com/questions/758107/solving-diophatine-equation-of-form-x2y2-25 – Martin Sleziak Apr 17 '14 at 18:49

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All Pythagorean triples, i.e. triplets of positive integers such that $a^2+b^2=c^2$, can be expressed as $a=k(m^2-n^2)$, $b=2kmn$, $c=k(m^2+n^2)$; where $m$ and $n$ are coprime, $m>n$ and $m$ and $n$ have opposite parity. (Up to interchanging of $a$ and $b$.)

So in your case you want $k(m^2+n^2)=25$. This gives you the possibilities:
$k=1$ and $m^2+n^2=25$
$k=5$ and $m^2+n^2=5$
$k=25$ and $m^2+n^2=1$

For such small numbers, it is easy to find all expressions as a sum of two squares by hand, you will get $25=0^2+5^2=3^2+4^2$, $5=1^2+2^2$ and $1=1^2+0^2$.

Since you are interested only in non-zero values, you get only two possibilities for $(k,m,n)$ namely $(1,4,3)$ and $(5,2,1)$. From the first one you get $a=7$, $b=24$, $c=25$. The second one gives you $a=15$, $b=20$, $c=25$.

Martin Sleziak
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