1. To find a (rational) right-angled triangle such that the hypotenuse minus each of the sides gives a cube.
Let the required triangle be formed from x, 3.
Therefore hypotenuse = x^2 + 9, perpendicular = 6x, base = x^2 – 9.
Thus x^2 + 9 – (x^2 - 9) = 18 should be a cube, but it is not.
Now 18 = 2.3^2; therefore we must replace 3 by m, where 2.m^2 is a cube; and m=2.
We form, therefore, a right-angled triangle from x, 2, namely (x^2 + 4, 4x, x^2 – 4); and one condition is satisfied.
The other gives x^2 – 4x + 4 = a cube;
therefore (x – 2)^2 is a cube, or x – 2 is a cube = 8, say.
Thus x = 10, and the triangle is (40, 96, 104).
2. To find a right-angled triangle such that the hypotenuse added to each side gives a cube.
Form a triangle, as before, from two numbers; and, as before, one of them must be such that twice its square is a cube, i.e. it must be 2.
We form a triangle from x, 2, namely x^2 + 4, 4x, 4 – x^2; therefore x^2 + 4x + 4 must be a cube, while x^2 must be less than 4, or x<2.
Thus x+2 = a cube which must be <4 and >2 = 27/8, say.
Therefore x = 11/8, and the triangle is (135/64, 11/2, 377/64), or, if we multiply by the common denominator, (135, 352, 377).
3. To find a right-angled triangle such that its area added to a given number makes a square.
Let 5 be the given number, (3x, 4x, 5x) the required triangle.
Therefore 6x^2 + 5 = a square = 9x^2, say,
or 3x^2 = 5.
But 3 should have to 5 the ratio of a square to a square.
Therefore we must find a right-angled triangle and a number such that the difference between the square of the number and the area of the triangle has to 5 the ratio of a square to a square, i.e. = 1/5 of a square.
Form a right-angled triangle from (m, 1/m);
thus the area is m^2 – 1/m^2.
Let the number be m + 2.5/m, so that we must have
4.5 + 101/m^2 = 1/5 of a square;
therefore 4.25 + 505/m^2 = a square,
or 100m^2 + 505 = a square = (10m + 5)^2, say,
and m = 24/5.
The auxiliary triangle must therefore be formed from 24/5, 5/24, and the auxiliary number sought is 413/60.
Put now for the original triangle (hx, px, bx), where (h, p, b) is the right-angled triangle formed from 24/5, 5/24;
this gives pbx^2/2 + 5 = 170569x^2/3600,
and we have the solution.
[The perpendicular sides of the right-angled triangle are
(24^2/5^2 – 5^2/24^2)x = 331151x/14400 and 2x,
whence 331151x^2/14400 + 5 = 170569x^2/3600,
x = 24/53,
and the triangle is (331151/31800, 48/53, 332401/31800).]
4. To find a right-angled triangle such that its area minus a given number makes a square.
Given number 6, triangle (3x, 4x, 5x), say.
Therefore 6x^2 – 6 = a square = 4x^2, say.
Thus, in this case, we must find a right-angled triangle and a number such that (area of triangle) – (number)^2 = 1/6 of a square.
Form a triangle from m, 1/m.
Its area is m^2 – 1/m^2, and let the number be m – (6/2).(1/m).
Therefore 6 – 10/m^2 = 1/6 (a square),
or 36m^2 – 60 = a square = (6m – 2)^2, say.
Therefore m = 8/3, and the auxiliary triangle is formed from (8/3, 3/8), the auxiliary number being 37/24.
We start again, substituting for 3, 4, 5 in the original hypothesis the sides of the auxiliary triangle just found, and putting (37/24)^2.x^2 in place of 4x^2; and the solution is obvious.
[The auxiliary triangle is (4015/576, 2, 4177/576), whence
4015x^2/576 – 6 = (37^2/24^2)x^2, and x = 8/7,
so that the required triangle is (4015/504, 16/7, 4177/504).]
5. To find a right-angled triangle such that, if its area be subtracted from a given number, the remainder is a square.
Given number 10, triangle (3x, 4x, 5x), say.
Thus 10 – 6x^2 = a sqaure; and we have to find a right-angled triangle and a number such that (area of triangle) + (number)^2 = 1/10 of a square.
Form a triangle from m, 1/m, the area being m^2 – 1/m^2, and let the number be 1/m + 5m.
Therefore 26m^2 + 10 = 1/10 of a square,
or 260m^2 + 100 = a square,
or again 65m^2 + 25 = a square = (8m + 5)^2, say,
whence m = 80.
The reat is obvious.
[The required triangle is (40959999/825600, 2/129, 40960001/825600).]
6. To find a right-angled triangle such that the area added to one of the perpendiculars makes a given number.
Given number 7, triangle(3x, 4x, 5x).
Therefore 6x^2 + 3x = 7.
In order that this might be solved, it would be necessary that (half coefficient of x)^2 + product of coefficient of x^2 and absolute term should be a square;
but (3/2)^2 + 6.7 is not a square.
Hence we must find, to replace (3, 4, 5), a right-angled triangle such that
(½ one perpendicular)^2 + 7 times area = a square.
Let one perpendicular be m, the other 1.
Therefore 7m/2 + ¼ = a square, or 14m + 1 = a square.
Also, since the triangle is rational, m^2 + 1 = a square.
The difference, m^2 – 14m = m(m – 14);
and putting, as usual, 7^2 = 14m + 1,
we have m = 24/7.
The auxiliary triangle is therefore (24/7, 1, 25/7) or (24, 7, 25).
Starting afresh, we take as the triangle (24x, 7x, 25x).
Therefore 84x^2 + 7x = 7,
and x = ¼.
We have then (6, 7/4, 25/4) as the solution.
7. To find a right-angled triangle such that its area minus one of the perpendiculars is a given number.
Given number 7.
As before, we have to find a right-angled triangle such that (½ one perpendicular)^2 + 7 times area = a square;
this triangle is (7, 24, 25).
Let then the triangle of the problem be (7x, 24x, 25x).
Therefore 84x^2 – 7x = 7,
x = 1/3,
and the problem is solved.
8. To find a right-angled triangle such that the area added to the sum of the perpendiculars makes a given number.
Given number 6.
Again I have to find a right-angled triangle such that
(½ sum of perpendiculars)^2 + 6 times area = a square.
Let m, 1 be the perpendicular sides of this triangle;
therefore ¼(m + 1)^2 + 3m = m^2/4 + 7m/2 + ¼ = a square,
while m^2 + 1 must also be a square.
Therefore m^2 + 14m +1 and m^2 +1 are both squares.
The difference is 2m.7, and we put
m^2 – 7m + 49/4 = m^2 +1,
whence m = 45/28,
and the auxiliary triangle is (45/28, 1, 53/28), or (45, 28, 53).
Assume now for the triangle of the problem (45x, 28x, 53x).
Therefore 630x^2 + 73x = 6;
x is rational [= 1/18], and the solution follows.
9. To find a right-angled triangle such that the area minus the sum of the perpendiculars is a given number.
Given number 6.
As before we find a subsidiary right-angled triangle such that (½ sum of perpendiculars)^2 + 6 times area = a square.
This is found to be (28, 45, 53) as before.
Taking (28x, 45x, 53x) for the required triangle,
630x^2 – 73x = 6;
x = 6/35, and the problem is solved.
10. To find a right-angled triangle such that the sum of its area, the hypotenuse, and one of the perpendiculars is a given number.
Given number 4.
If we assumed as the triangle (kx, px, bx), we should have
pbx^2/2 + hx + bx = 4;
and, in order that the solution may be rational, we must find a right-angled triangle such that
¼(hyp. + one perp.)^2 + 4 times area = a square.
Form a right-angled triangle from 1, m+1.
Then ¼(hyp. + one perp.)^2 = ¼(m^2 + 2m + 2 + m^2 + 2m)^2
= m^4 + 4m^3 + 6m^2 + 4m + 1,
and 4 times area = 4(m+1)(m^2 + 2m)
= 4m^3 + 12m^2 + 8m.
Therefore
m^4 + 8m^3 + 18m^2 + 12m + 1 = a square = (6m + 1 – m^2)^2, say, whence m = 4/5, and the auxiliary triangle is formed from (1, 9/5) or (5, 9). This triangle is (56, 90, 106) or (28, 45, 53).
We therefore assume 28x, 45x, 53x for the original triangle,
and we have 630x^2 + 81x = 4.
Therefore x = 4/105, and the problem is solved.
11. To find a right-angled triangle such that its area minus the sum of the hypotenuse and one of the perpendiculars is a given number.
Given number 4.
We have then to find an auxiliary triangle with the same property as in the last problem;
therefore (28, 45, 53) will serve the purpose.
We put for the triangle of the problem (28x, 45x, 53x), and we have
630x^2 – 81x = 4;
x = 1/6, and the problem is solved.
Lemma I to the following problem.
To find a right-angled triangle such that the difference of the perpendiculars is a square, the greater alone is a square, and further the area added to the lesser perpendicular gives a square.
Let the triangle be formed from two numbers, the greater perpendicular being twice their product.
Hence I must find two numbers such that (1) twice their product is a square and (2) twice their product exceeds the difference of their squares by a square.
This is true of any two numbers the greater of which = twice the lesser.
Form then the triangle from x, 2x, and two conditions are satisfied.
The third gives 6x^4 + 3x^2 = a square, or 6x^2 + 3 = a square.
I have therefore to find a number such that 6 times its square + 3 = a square;
one such number is 1, and there are an infinite number of others.
If x = 1, the triangle is formed from 1, 2.
Lemma II to the following problem.
Given two numbers the sum of which is a square, an infinite number of squares can be found such that, when the square is multiplied by one of the given numbers and the product is added to the other, the result is a square.
Given numbers 3, 6.
Let x^2 + 2x + 1 be the required square which, say, when multiplied by 3 and then increased by 6, gives a square.
We have 3x^2 + 6x + 9 = a square;
and, since the absolute term is a square, an infinite number of solutions can be found.
Suppose, e.g., 3x^2 + 6x + 9 = (3 – 3x)^2,
and x = 4.
The side of the required square is 5, and an infinite number of other solutions can be found.
12. To find a right-angled triangle such that the area added to either of the perpendiculars gives a square.
Let the triangle be (5x, 12x, 13x).
Therefore (1) 30x^2 + 12x = a square = 36x^2, say,
and x = 2.
But (2) we must also have
30x^2 + 5x = a square.
This is however not a square when x = 2.
Therefore I must find a square m^2x^2, to replace 36x^2, such that 12/(m^2 – 30), the value of x obtained from the first equation, is real and satisfies the condition
30x^2 + 5x = a square.
This gives, by substitution,
(60m^2 + 2520)/(m^2 – 60m + 900) = a square,
or 60m^2 + 2520 = a square.
This could be solved [by the preceding Lemma II] if 60 + 2520 were equal to a square.
Now 60 arises from 5.12, i.e. from the product of the perpendicular sides of (5, 12, 13);
2520 is 30.12.(12 – 5), i.e. the continued product of the area, the greater perpendicular, and the difference between the perpendiculars.
Hence we must find an auxiliary triangle such that (product of perps.) + (continued product of area, greater perp. and difference of perps.) = a square.
Or, if we make the greater perpendicular a square, and divide out by it, we must have
(lesser perp.) + (product of area and diff. of perps.) = a square.
Then, assuming that we have found two numbers, (1) the product of the area and the difference of the perpendiculars and (2) the lesser perpendicular, satisfying these conditions, we have to find a square (m^2) such that the product of this square into the second of the numbers, when added to the first number, gives a square.
How to solve these problems is shown in the Lemmas.
The auxiliary triangle is (3, 4, 5). [Lemma I]
Accordingly, putting for the original triangle (3x, 4x, 5x),
we have 6x^2 + 4x, 6x^2 + 3x, both squares.
Let x = 4/(m^2 – 6) be the solution of the first equation;
then x^2 = 16/(m^4 – 12m^2 + 36).
The second equation therefore gives 96/(m^4 – 12m^2 + 36) + 12/(m^2 – 6) = a square,
whence 12m^2 + 24 = a square,
and we have therefore to find a square (m^2) such that twelve times it + 24 is a square; this is possible, since 12 + 24 is a square [Lemma II].
A solution is m^2 = 25,
whence x = 4/19, and (12/19, 16/19, 20/19) is the required triangle.
13. To find a right-angled triangle such that its area minus either perpendicular gives a square.
We have to find an auxiliary triangle exactly as in the last problem;
this triangle is (3, 4, 5), and accordingly we assume for the triangle of the problem (3x, 4x, 5x).
One condition then gives 6x^2 – 4x = a square = m^2x^2, say (m^2 < 6),
and x = 4/(6 – m^2).
The second condition gives 6x^2 – 3x = a square; and, by substitution,
96/(m^4 – 12m^2 + 36) – 12/(6 – m^2) = a square,
or 24 + 12m^2 = a square.
This is satisfied by m = 1, whence x = 4/5, and the required triangle is (12/5, 16/5, 4).
Or, if we do not wish to use the value 1 for m, let m = z + 1, and (dividing by 4) we have
3m^2 + 6 = 3z^2 + 6z + 9 = a square;
z must be found to be not greater than 13/9 (in order that m^2 may be less than 6), and m will not be greater than 22/9. The solution is then rational.
14. To find a right-angled triangle such that its area minus the hypotenuse or minus one of the perpendiculars gives a square.
Let the triangle be (3x, 4x, 5x).
Therefore 6x^2 – 5x, and 6x^2 – 3x are both squares.
Making the latter a square (= m^2x^2), we have
x = 3/(6 – m^2), [m^2 < 6].
The first equation then gives 54/(m^4 – 12m^2 + 36) – 15/(6 – m^2) = a square,
or 15m^2 – 36 = a square.
This equation we cannot solve because 15 is not the sum of two squares.
Therefore we must change the assumed triangle.
Now (with reference to the triangle 3, 4, 5) 15m^2 = the continued product of a square less than the area, the hypotenuse, and one perpendicular;
while 36 = the continued product of the area, the perpendicular, and the difference between the hypotenuse and the perpendicular.
Therefore we have to find a right-angled triangle (h, p, b, say) and a square (m^2) less than 6 such that m^2hp – (pb/2).p(h – p) is a square.
If we form the triangle from two numbers X1, X2 and suppose that p = 2X1X2, and if we then divide throughout by (X1 – X2)^2 which is equal to h – p, we must find a square z^2 [= m^2/(X1 – X2)^2] such that z^2hp – (pb/2).p is a square.
The problem can be solved if X1, X2 are "similar plane numbers".
Form the auxiliary triangle from similar plane numbers accordingly, say 4, 1. [The conditions are then satisfied.]
[The equation for m then becomes
8.17m^2 – 4.15.8.9 = a square,
or 136m^2 – 4320 = a square.]
Let m^2 = 36. [This satisfies the equation, and 36 < area of triangle.]
The triangle formed from 4, 1 being (8, 15, 17), we assume 8x, 15x, 17x for the original triangle.
We now put 60x^2 – 8x = 36x^2,
and x = 1/3.
The required triangle is therefore (8/3, 5, 17/3).
Lemma to the following problem.
Given two numbers, if, when some square is multiplied into one of the numbers and the other number is subtracted from the product, the result is a square, another square larger than the aforesaid square can always be found which has the same property.
Given numbers 3, 11, side of square 5, say, so that
3.25 – 11 = 64, a square.
Let the required square be (x + 5)^2.
Therefore 3(x + 5)^2 – 11 = 3x^2 + 30x + 64 = a square, = (8 – 2x)^2, say,
and x = 62.
The side of the new square is 67, and the square itself 4489.
15. To find a right-angled triangle such that the area added to either the hypotenuse or one of the perpendiculars gives a square.
In order to guide us to a proper assumption for the required triangle, we have, in this case, to seek a triangle (h, p, b, say) and a square (m^2) such that m^2 > pb/2, the area, and
m^2hp – (pb/2).p(h – p) is a square.
Let the triangle be formed from 4, 1, the square (m^2) being 36, as before;
but the triangle being (8, 15, 17), the square is not greater than the area.
We must therefore, as in the preceding Lemma, replace 36 by a greater square.
Now hp = 136, and (pb/2).p.(h – p) = 60.8.9 = 4320,
so that 136m^2 – 4320 = a square,
which is satisfied by m^2 = 36; and we have to find a larger square (z^2) such that
136z^2 – 4320 = a square.
Put z = m + 6, and we have
(m^2 + 12m + 36)136 – 4320 = a square,
or 136m^2 + 1632m + 576 = a square = (km – 24)^2, say.
This equation has any number of solutions; e.g. putting k = 16, we have
m = 20, z = 26, and z^2 = 676.
We therefore put (8x, 15x, 17x) for the original triangle, and then assume
60x^2 + 8x = 676x^2,
whence x = 1/77, and the problem is solved.
16. To find a right-angled triangle such that the number representing the (portion intercepted within the triangle of the) bisector of an acute angle is rational.
Suppose the bisector AD = 5x, and one segment of the base (DB) = 3x; therefore the perpendicular = 4x.
Let the whole base CB be some multiple of 3, say 3; then CD = 3 – 3x.
But since AD bisects the angle CAB,
AC:CD = AB:BD;
therefore the hypotenuse AC = (4/3)(3 – 3x) = 4 – 4x.
Therefore [by Eucl. I. 47]
16x^2 – 32x + 16 = 16x^2 + 9,
and x = 7/32.
If we multiply throughout by 32, the perpendicular = 28, the base = 96, the hypotenuse = 100, and the bisector = 35.
17. To find a right-angled triangle such that the area added to the hypotenuse gives a square, while the perimeter is a cube.
Let the area be x and the hypotenuse some square minus x, say 16 – x.
The product of the perpendiculars = 2x;
therefore, if one of them be 2, the other is x, and the perimeter = 18, which is not a cube.
Therefore we must find some square which, when 2 is added to it, becomes a cube.
Let the side of the square be m+1, and that of the cube m-1.
Therefore m^3 – 3m^2 + 3m – 1 = m^2 + 2m + 3,
from which m is found to be 4.
Hence the side of the square = 5, and that of the cube = 3.
Assuming now x for the area of the original triangle, 25 – x for its hypotenuse, and 2, x for the perpendiculars, we find that the perimeter is a cube.
But (hypotenuse)^2 = sum of squares of perpendiculars; therefore
x^2 – 50x + 625 = x^2 + 4;
x = 621/50, and the problem is solved.
18. To find a right-angled triangle such that the area added to the hypotenuse gives a cube, while the perimeter is a square.
Area x, hypotenuse some cube minus x, perpendiculars x, 2.
Therefore we have to find a cube which when 2 is added to it, becomes a square.
Let the side of the cube be m – 1.
Therefore m^3 – 3m^2 + 3m + 1 = a square = (3m/2 + 1)^2, say.
Thus m = 21/4, and the cube = (17/4)^3 = 4913/64.
Put now x for the area, x, 2 for the perpendiculars, and 4913/64 – x for the hypotenuse;
and x is found from the equation (4913/64 – x)^2 = x^2 + 4.
[x = 24121185/628864, and the triangle is (2, 24121185/628864, 24153953/628864).]
19. To find a right-angled triangle such that its area added to one of the perpendiculars gives a square, while the perimeter is a cube.
Make a right-angled triangle from some indeterminate odd number, say 2x + 1;
then the altitude = 2x + 1, the base = 2x^2 + 2x, and the hypotenuse = 2x^2 + 2x + 1.
Since the perimeter = a cube,
4x^2 + 6x + 2 = (4x + 2)(x + 1) = a cube;
and, if we divide all the sides by x + 1, we have to make 4x + 2 a cube.
Again, the area + one perpendicular = a square.
Therefore (2x^3 + 3x^2 + x)/(x + 1)^2 + (2x + 1)/(x + 1) = a square;
that is, (2x^3 + 5x^2 + 4x + 1)/(x^2 + 2x + 1) = 2x + 1 = a square.
But 4x + 2 = a cube;
therefore we must find a cube which is double of a square; this is of course 8.
Therefore 4x + 2 = 8, and x = 3/2.
The required triangle is (8/5, 3, 17/5).
20. To find a right-angled triangle such that the sum of its area and one perpendicular is a cube, while its perimeter is a square.
Proceeding as in the last problem, we have to make 4x + 2 a square, 2x + 1 a cube.
We have therefore to seek a square which is double of a cube; this is 16, which is double of 8.
Therefore 4x + 2 = 16, and x = 7/2.
The triangle is (16/9, 7, 65/9).
21. To find a right-angled triangle such that its perimeter is a square, while its perimeter added to its area gives a cube.
Form a right-angled triangle from x, 1.
The perpendiculars are then 2x, x^2 – 1, and the hypotenuse x^2 + 1.
Hence 2x^2 + 2x should be a square, and x^3 + 2x^2 + x a cube.
It is easy to make 2x^2 + 2x a square; let 2x^2 + 2x = m^2x^2; therefore x = 2/(m^2 – 2).
By the second condition,
8/(m^2 – 2)^3 + 8/(m^2 - 2 )^2 + 2/(m^2 – 2) must be a cube,
i.e. 2m^4/(m^2 - 2)^3 = a cube.
Therefore 2m^4 = a cube, or 2m = a cube = 8, say.
Thus m = 4, x = 2/14 = 1/7, x^2 = 1/49.
But one of the perpendiculars of the triangle is x^2 – 1, and we cannot subtract 1 from 1/49.
Therefore we must find another value for x greater than 1; hence
2 < x^2 < 4.
And we have therefore to find a cube such that ¼ of the square of it is greater than 2, but less than 4.
If z^2 be this cube,
2 < z^6/4 < 4,
or 8 < z^6 < 16.
This is satisfied by z^6 = 729/64, or z^3 = 27/8.
Therefore m = 27/16, m^2 = 729/256, and x = 512/217, the square of which is > 1.
Thus the triangle is known [1024/217, 215055/47089, 309233/47089].
22. To find a right-angled triangle such that its perimeter is a cube, while the perimeter added to the area gives a square.
(1) We must first see how, given two numbers, a triangle may be formed such that its perimeter = one of the numbers and its area = the other.
Let 12, 7 be the numbers, 12 being the perimeter, 7 the area.
Therefore the product of the two perpendiculars = 14 = (1/x).14x.
If then 1/x, 14x are the perpendiculars,
hypotenuse = perimeter – sum of perps. = 12 – 1/x – 14x.
Therefore [by Eucl. I. 47]
1/x^2 + 196x^2 + 172 – 24/x – 336x = 1/x^2 + 196x^2;
that is,
172 = 336x + 24/x,
or 172x = 336x^2 + 24.
This equation gives no rational solution, because 86^2 – 24.336 is not a square.
Now 172 = (perimeter)^2 + 4 times area,
24.336 = 8 times area multiplied by (perimeter)^2.
(2) Let now the area = m, and the perimeter = any number which is both a square and a cube, say 64.
Therefore ((64^2 + 4m)/2)^2 – 8.64^2.m must be a square,
or 4m^2 – 24576m + 4194304 = a square.
Therefore m^2 – 6144m + 1048576 = a square, and also m + 64 = a square.
To solve this double-equation, multiply the second by such a number as will make the absolute term the same as the absolute term in the first.
Then, if we take the difference and the factors as usual, the equations are solved.
[After the second equation is multiplied by 16384, the double equation becomes
m^2 – 6144m + 1048576 = a square and
16384m + 1048576 = a square.
The difference is m^2 – 22528m.
If m, m – 22528 are taken as the factors, we find m = 7680, which is an impossible value for the area of a right-angled triangle of perimeter 64.
We therefore take as the factors 11m, m/11 – 2048; then, when the square of half the difference is equated to the smaller of the two expressions to be made squares, we have
(60m/11 + 1024)^2 = 16384m + 1048576,
and m = 39424/225.
Returning now to the original problem, we put 1/x, 2mx for the perpendicular sides of the required triangle, and we have
(64 – 1/x – 2mx)^2 = 1/x^2 + 4m^2x^2,
which leads, when the value of m is substituted, to the equation
78848x^2 – 8432x + 225 = 0.
The solution of this equation is rational, namely
x = (527 +/- 23)/9856 = 25/448 or 9/176.
The required triangle is thus (448/25, 176/9, 5968/225).
23. To find a right-angled triangle such that the square of its hypotenuse is also the sum of a different square and the side of that square, while the quotient obtained by dividing the square of the hypotenuse by one of the perpendiculars of the triangle is the sum of a cube and the side of the cube.
Let one of the perpendiculars be x, the other x^2.
Therefore (hypotenuse)^2 = the sum of a square and its side; also (x^4 + x^2)/x = x^3 + x = the sum of a cube and its side.
It remains that x^4 + x^2 must be a square.
Therefore x^2 + 1 = a square = (x – 2)^2, say.
Therefore x = ¾, and the triangle is found [¾, 9/16, 15/16].
24. To find a right-angled triangle such that one perpendicular is a cube, the other is the difference between a cube and its side, and the hypotenuse is the sum of a cube and its side.
Let the hypotenuse be x^3 + x, and one perpendicular x^3 – x.
Therefore the other perpendicular = 2x^2 = a cube = x^3, say.
Thus x = 2, and the triangle is (6, 8, 10).