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# Help!

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5) (n, k) are positive integers, solve n(n + 2) = k^2

6) (n, k) are positive integers, solve n(n + 4) = k^2

7) (n, k) are positive integers, solve n(n + 5) = k^2

Feb 4, 2021

#1
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5) (n, k) are positive integers, solve n(n + 2) = k^2
6) (n, k) are positive integers, solve n(n + 4) = k^2

Hello Guest!

$$\{n,k\}\subset \mathbb Z$$

$$n(n+2)=k^2$$

$$k\in\{$$ 4, 9, 16, 25, 36, 49, 64, 81, 100, 121   $$\}$$

$$n\in\{$$ 1, 2,  3,   4,    5,  6,   7,   8,   9,   10,   11,   12$$\}$$

$$n(n+2)\in \{$$3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168$$\}$$

No solution for positive n, but if n = -1

$$-1(-1+2)=i^2$$

$$n(n+5)=k^2$$

No solution for positive n, but if n = -4

$$-4(-4+5)=(2i)^2$$

!

Feb 4, 2021
#2
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Hello other guest! The answer to 5) is no solution.

n(n+2)=k^2

n^2+2n+1=k^2+1

(n+1)^2=k^2+1

(n+1)^2-k^2=1

(n+1+k)*(n+1-k)=1

Now, in order for this to be equal, n and k have to equal 0 but you said they must be positive integers so I believe that there is NO SOLUTION.

Feb 4, 2021