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- Homework Statement
- Determine (with proof) the set of all prime numbers that can divide two successive integers of the form ## n^{2}+3 ##.

- Relevant Equations
- None.

Proof:

Let ## p ## be the prime divisor of two successive integers ## n^{2}+3 ## and ## (n+1)^{2}+3 ##.

Then ## p\mid [(n+1)^{2}+3-(n^{2}+3)]\implies p\mid (2n+1) ##.

Observe that ## p\mid (n^{2}+3) ## and ## p\mid (2n+1) ##.

Now we see that ## p\mid [(n^{2}+3)-3(2n+1)]\implies p\mid (n^{2}-6n)\implies p\mid [n(n-6)] ##.

Since ## p\nmid n ##, it follows that ## p\mid (n-6) ##.

By Euclidean Algorithm, we have that ## p\mid (2n+1)\implies p\mid [2(n-6)+13] ##.

Therefore, the set of all prime numbers that can divide two successive integers of the form ## n^{2}+3 ## is ## {13} ##.

Let ## p ## be the prime divisor of two successive integers ## n^{2}+3 ## and ## (n+1)^{2}+3 ##.

Then ## p\mid [(n+1)^{2}+3-(n^{2}+3)]\implies p\mid (2n+1) ##.

Observe that ## p\mid (n^{2}+3) ## and ## p\mid (2n+1) ##.

Now we see that ## p\mid [(n^{2}+3)-3(2n+1)]\implies p\mid (n^{2}-6n)\implies p\mid [n(n-6)] ##.

Since ## p\nmid n ##, it follows that ## p\mid (n-6) ##.

By Euclidean Algorithm, we have that ## p\mid (2n+1)\implies p\mid [2(n-6)+13] ##.

Therefore, the set of all prime numbers that can divide two successive integers of the form ## n^{2}+3 ## is ## {13} ##.