Question

if x and y are prime numbers which satisfy x^2-2y^2=1, solve for x and y.​

Answer 1

Solution :-

→ x² - 2y² = 1

→ x² = (1 + 2y²)

Now, as we can see , (1 + 2y²) is always a odd number ,

Therefore, x² is also a odd number .

Or,

we can say that, x is a odd number .

Let us Assume that, x = (2k + 1) . { for all values of k.}

So,

→ (2k + 1)² = (1 + 2y²)

using (a + b)² = a² + b² + 2ab in LHS,

→ 4k² + 4k + 1 = (1 + 2y²)

→ 4k² + 4k = 2y²

→ 2(2k² + 2k) = 2y²

→ 2k² + 2k = y²

→ 2(k² + k) = y²

conclude that, y² or y will also be a Even number , as it is multiple of 2.

Now, we have given that, y is a prime number . And, we know that, only Even Prime number is 2.

Therefore,

→ y = 2 .

Hence ,

→ x² = (1 + 2y²)

→ x² = 1 + 2(2)²

→ x² = 1 + 8

→ x² = 9

→ x = 3.

∴ only value of (x,y) will be (3,2).

Answer 2

Answer:

x = 3, y = 2

Step-by-step explanation:

SOLUTION:-

Given that,

=> x² - 2y² = 1

=> x² = 2y² + 1 .........(i)

From (i), we can conclude that x is an odd number.

So,

Let x = 2k + 1

{ As we know that any odd integer is of form 2k + 1 }

=> x² = (2k + 1)²

=> x² = 4k² + 4k + 1 .........(ii)

From (i) and (ii), we have,

=> 2y² + 1 = 4k² + 4k + 1

=> 2y² = 4k² + 4k

=> y² = 2k² + 2k

So, y is an even number.

But it is also specified that y is a prime number.

So, the only even prime number is 2.

Thus,

★[ y = 2 ]★

Put y = 2 in (i)

=> x² = 2(2)²+1

=> x² = 8+1 = 9

=> x = √9

★[ x = 3 ]★

Hence, the value of x and y are 3 and 2 respectively.

ie. (x,y) = (3,2)

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