Right angled triangles with sides a and b and hypotenuse c obey Pythagoras' Theorem

a^{2} + b^{2} = c^{2}

Suppose a, b and c are integers with no common factor. Then both a and b
cannot be even, since, if they were, c would be even and 2 would be a
common factor. Again, both a and b cannot be odd, since any odd square
has form 4*n+1 and the sum of 2 odd numbers has form 8*n+2. This is
divisible by 2 but not 4, and thus cannot be a square.

Thus, one side must be even and the other odd and c must odd. Call the
even one b.

then b^{2} = c^{2} -a^{2} = (c-a)*(c+a)

c-a and c+a are unequal factors of a square number and hence must
both be squares. Furthermore, they are both even. Thus let

4*u^{2} = c+a and 4*v^{2} = c-a.

Hence

b = 2*u*v and a = u^{2} - v^{2} and
c = u^{2} + v^{2}

Similar arguments show that u > v, u and v have no common factor,
and they cannot both be odd.

The area of a right angled triangle is a*b/2 = u*v*(u^{2}-v^{2})

If the sides have a common factor k then the area is

k^{2}*a*b/2 = k^{2}*u*v*(u^{2}-v^{2})

My program used this formula to generate the areas of all integer-sided
right
angled triangles by letting k and u be any integers and v be any integer
less than u
with no common factor and u and v not both odd. It used values of u
from 2 to 100
and then multiplied by k^{2} while the area was less than 1 million.
Each triple of (area, a, b) was placed in a file and, when complete, the
file
was sorted in order of increasing area. This brought together triangles
with the
same area and a simple search program found the first group of 4 of the same
area.

Their sides are

a | b | c |
---|---|---|

518 | 1320 | 1418 |

280 | 2442 | 2458 |

231 | 2960 | 2969 |

111 | 6160 | 6161 |

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