The side lengths of a right triangle are pairwise relatively prime integers. By increasing the lengths of the legs by 100100 and the hypotenuse by 140 units, we again obtain the side lengths of a right triangle. Determine the sides of the two triangles.
Let the legs of the first triangle be $a$ and $b$, and the hypotenuse $c$, then for the second triangle, the legs are $a+100, b+100$ and the hypotenuse is $c+140$, and according to the Pythagorean theorem,

$$
a^{2}+b^{2}=c^{2} \quad \text { and } \quad(a+100)^{2}+(b+100)^{2}=(c+140)^{2}
$$

Subtracting the first from the second and rearranging and simplifying,

$$
5(a+b+2)=7 c
$$

This can only hold if $c$ is divisible by 5: $c=5 c^{\prime}$, where $c^{\prime}$ is also an integer. Then

$$
a+b=7 c^{\prime}-2
$$

Using this, from our first equation we can also express $a b$ in terms of $c^{\prime}$:

$$
25 c^{\prime 2}=a^{2}+b^{2}=(a+b)^{2}-2 a b=\left(7 c^{\prime}-2\right)^{2}-2 a b
$$

Rearranging and simplifying,

$$
a b=12 c^{\prime 2}-14 c^{\prime}+2
$$

Based on these, $a$ and $b$ are the two roots of the quadratic equation in $x$:

$$
x^{2}-\left(7 c^{\prime}-2\right) x+\left(12 c^{\prime 2}-14 c^{\prime}+2\right)=0
$$

These can only be integers if the discriminant is the square of an integer $t$:

$$
\left(7 c^{\prime}-2\right)^{2}-4\left(12 c^{\prime 2}-14 c^{\prime}+2\right)=c^{\prime 2}+28 c^{\prime}-4=\left(c^{\prime}+14\right)^{2}-200=t^{2}
$$

From this,

$$
\left(c^{\prime}+14\right)^{2}-t^{2}=\left(c^{\prime}+t+14\right)\left(c^{\prime}-t+14\right)=200
$$

The factors of the product have the same parity, because they differ by $2 t$, which is an even number. Thus, both are even, since their product is even. We can assume that $t$ is positive, then the first factor is positive, and so is the second, and the first is the larger. The possible factorizations of 200 are: $200=100 \cdot 2=50 \cdot 4=20 \cdot 10$. Denoting the two factors by $2 m$ and $2 n$,

$$
c^{\prime}+t+14=2 m, \quad c^{\prime}-t+14=2 n
$$

from which

$$
c^{\prime}=m+n-14, \quad t=m-n, \quad c=5 c^{\prime}=5 m+5 n-70
$$

and the legs, as the two roots of (2),

$$
a, b=\frac{7 c^{\prime}-2 \pm t}{2}=4 m+3 n-50, \quad \text { or } \quad 3 m+4 n-50
$$

Substituting the obtained pairs of values for $m, n$: $50,1 ; 25,2 ; 10,5$, we get the following triangle sides:

| $m$ | 50 | 25 | 10 |
| :--- | ---: | ---: | ---: |
| $n$ | 1 | 2 | 5 |
| $a$ | 153 | 56 | 5 |
| $b$ | 104 | 33 | 0 |
| $c$ | 185 | 63 | 5 |

The last triplet does not form the sides of a triangle, and we can verify that the first two satisfy the Pythagorean equation and the increased side lengths as per the problem statement. (This also holds for the third triplet.)

Remark. Similar reasoning leads us to the goal when we use the fact that if the side lengths of a right triangle are relatively prime integers, then they can be expressed in the form $a=u^{2}-v^{2}, b=2 u v, c=u^{2}+v^{2}$, where $u$ and $v$ are relatively prime natural numbers, one of which is even and the other odd, and we substitute these into (1).[^0]

[^0]:    ${ }^{1}$ See, for example, H. Rademacher - O. Toeplitz: Numbers and Figures, Secondary School Special Booklets, 2nd edition, Tankönyvkiadó, Budapest, $1954,87-88$. о.