True or False!!!!!!!!!!!!!?

There exist an x,y,z integer, such that x>37, y>37 z>37, and x^2 = y^2 + z^2.

Comments

  • True.

    x = 100, y = 80, z = 60

  • Do you mean "three integers x, y, z"? If so, then true. In fact, there are an infinite number of them.

    x^2 = y^2 + z^2 is the Pythagorean formula. Three integers that fit are x = 5, y = 3, z = 4

    5^2 = 3^2 + 4^2 = 25

    Multiply all by 13 so that all numbers are >37 and you still have the same relationship. In fact, any multiplier will keep the Pythagorean relationship, but it has to be >12 1/3 to ensure y > 37. Try it!

  • Why, of course. Your formula looks a lot like the Pythagorean Theorem, so why not use a Pythagorean triple and scale it by any factor that makes all three integers greater than 37?

    For example, 3-4-5 is a widely-recognized Pythagorean triple. Multiply that by, say, 100, and you have 300, 400, and 500. Say x = 500, y = 400, z = 300. Then x^2 = y^2 + z^2.

  • Yes. Ex x=5, y=3, z= 4 >>3^2+4^2=5^2 and all are >37 25=9+16

  • start with 3, 4, 5 = first Pythagorean triplet

    3^2 + 4^2 = 5^2

    multiply each by 13

    3 => 3 * 13

    4 => 4 * 13

    5 => 5 * 13

    x, y and z are now > 37 each

    and (13 * 3)^2 + (13 * 4)^2 = (13 *5)^2

    so there is at least one set of x,y,z that fit the criteria, so true.

  • True.

    For example, 100, 60, 80

  • True

    Use any Pythagorean triple (e.g. 3n,4n,5n) where n>= 13

    39^2+52^2 = 65^2

  • There are infinitely many Pythagorean triples. True. Example: 500² = 300² + 400²

  • True, pick any values for y and z that are greater than 37 and then solve for x.

  • ....Aint nobody got time fo math.

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