A Radical Equation | Integer Solutions

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[solomonwilliams]

You missed a couple at the ends. Zero is an integer.

[MichaelRothwell]

Wyboo's method easily extends to solving in integers any equation of the form
√x+√y=√n, where n is a non-negative integer.

We note that x, y both lie between 0 and n.

Now we write n=p²q where q is the product of the prime factors of n that occur an odd number of times in its prime factorisation, so that p² is its largest perfect square factor (p=7, q=2 in the case in the video).

√x+√y=√n=√(p²q)=p√q
⇔√y=p√q-√x ...(1)
⇒y=(p√q-√x)² (squaring both sides)
⇔y=p²q-2p√q√x+x
⇔√(4p²qx)=n+x-y
So 4p²qx=(2p)²qx is a perfect square, hence so is qx, so qx=k², where k is a non-negative integer.

As q is the product of distinct primes, k is also divisible by each of these primes, and so is divisible by q. Hence k=qa, where a is a non-negative integer, and we have

qx=(qa)²=q²a², x=qa² with 0≤qa²≤n=qp², 0≤a²≤p², 0≤a≤p (as a, p non-negative).

Also √x=a√q, and substituting in (1) we get
√y=p√q-a√q=(7-a)√q
y=q(7-a)²

We can also write the solutions in symmetric form as
x=qa², y=qb² where a, b integers with a+b=p and 0≤a≤p, 0≤b≤p, giving a total of p+1 solutions.

[copernic]

Since 7√2 = t√2 + (7-t)√2
we can parametrize x and y:
x = 2t²
y = 2(7-t)²
with 0 < t < 7.

[moeberry]

Syber come on you made a very simple error. You also have 0, 98 and 98, 0

[wyboo]

how i solved it:

solving for y we get to:

y = 98 + x - sqrt(392 x)

392 = 2^3*7^2

so:

y = 98 + x - sqrt(2^3 * 7^2 * x)

the expression under the radical is a perfect square when all of the powers are even, so the complete solution set is:

x = 2^(2a+1) * 7^(2b) * 3^(2c) * 5^(2d) * ...

where each term in the product is a power of a prime number, and a, b, c, d, and etc. are all integers

each of these solutions for different choices of integers for the powers is unique because prime factorizations are unique

[Bill-rkjl]

Well now my

whole day's going to be screwed up too much thinking it's early in the morning

[mircoceccarelli]

( 0, 98 ), ( 98, 0 ) sono altre due soluzioni !