Prove that n^{4}+4 is never a prime number for n>1.

84 thoughts on “Sunday number theory problem”

1) (n^4)+4 is always an even number (greater than 2) for all values of n.
2) Even numbers (greater than 2) are never prime.
3) Er… that’s it.

Or am I missing something?

Er, yes. n^{4} is odd if n is odd; so is n^{4} +4 !!!

Oops. Well, I’m only an accountant, not a mathematician.

1) (n^4)+4 is always an even number (greater than 2) for all values of n.
2) Even numbers (greater than 2) are never prime.
3) Er… that’s it.

Or am I missing something?

Er, yes. n^{4} is odd if n is odd; so is n^{4} +4 !!!

Oops. Well, I’m only an accountant, not a mathematician.

1) (n^4)+4 is always an even number (greater than 2) for all values of n.
2) Even numbers (greater than 2) are never prime.
3) Er… that’s it.

Or am I missing something?

Er, yes. n^{4} is odd if n is odd; so is n^{4} +4 !!!

Oops. Well, I’m only an accountant, not a mathematician.

1) (n^4)+4 is always an even number (greater than 2) for all values of n.
2) Even numbers (greater than 2) are never prime.
3) Er… that’s it.

Or am I missing something?

Er, yes. n^{4} is odd if n is odd; so is n^{4} +4 !!!

Oops. Well, I’m only an accountant, not a mathematician.

1) (n^4)+4 is always an even number (greater than 2) for all values of n.
2) Even numbers (greater than 2) are never prime.
3) Er… that’s it.

Or am I missing something?

Er, yes. n^{4} is odd if n is odd; so is n^{4} +4 !!!

Oops. Well, I’m only an accountant, not a mathematician.

1) (n^4)+4 is always an even number (greater than 2) for all values of n.
2) Even numbers (greater than 2) are never prime.
3) Er… that’s it.

Or am I missing something?

Er, yes. n^{4} is odd if n is odd; so is n^{4} +4 !!!

Oops. Well, I’m only an accountant, not a mathematician.

Spoilers ahead!

Suppose n ≠ 0 (modulo 5). Then n^{4} = 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) So n^{4} + 4 is divisible by 5 (and so can only be prime if n = 1).

Suppose on the other hand that n = 0 (modulo 5). If n is even, then so is n^{4}, and so is n^{4} + 4.

So suppose that n is an odd multiple of 5, that is, n = 5(2a + 1) for some a ≥ 0. Then

So n^{4} + 4 cannot be prime in this case either. (For example, 5^{4} + 4 = 17 × 37; and 15^{4} + 4 = 197 × 257.)

I managed the first two steps of this myself; but in fact neither is necessary, there is a perfectly decent proof for all n>1, which your third step comes close to getting.

Ah yes, I see, you can find a similar product for all odd n, not just odd multiples of 5.

Indeed, for all n odd or even (though of course the even ns are trivially proved anyway).

Just to put you out of your misery (if you were in that state):

For n=1, ((n-1)^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.

Spoilers ahead!

Suppose n ≠ 0 (modulo 5). Then n^{4} = 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) So n^{4} + 4 is divisible by 5 (and so can only be prime if n = 1).

Suppose on the other hand that n = 0 (modulo 5). If n is even, then so is n^{4}, and so is n^{4} + 4.

So suppose that n is an odd multiple of 5, that is, n = 5(2a + 1) for some a ≥ 0. Then

So n^{4} + 4 cannot be prime in this case either. (For example, 5^{4} + 4 = 17 × 37; and 15^{4} + 4 = 197 × 257.)

I managed the first two steps of this myself; but in fact neither is necessary, there is a perfectly decent proof for all n>1, which your third step comes close to getting.

Ah yes, I see, you can find a similar product for all odd n, not just odd multiples of 5.

Indeed, for all n odd or even (though of course the even ns are trivially proved anyway).

Just to put you out of your misery (if you were in that state):

For n=1, ((n-1)^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.

Spoilers ahead!

Suppose n ≠ 0 (modulo 5). Then n^{4} = 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) So n^{4} + 4 is divisible by 5 (and so can only be prime if n = 1).

Suppose on the other hand that n = 0 (modulo 5). If n is even, then so is n^{4}, and so is n^{4} + 4.

So suppose that n is an odd multiple of 5, that is, n = 5(2a + 1) for some a ≥ 0. Then

So n^{4} + 4 cannot be prime in this case either. (For example, 5^{4} + 4 = 17 × 37; and 15^{4} + 4 = 197 × 257.)

I managed the first two steps of this myself; but in fact neither is necessary, there is a perfectly decent proof for all n>1, which your third step comes close to getting.

Ah yes, I see, you can find a similar product for all odd n, not just odd multiples of 5.

Indeed, for all n odd or even (though of course the even ns are trivially proved anyway).

Just to put you out of your misery (if you were in that state):

For n=1, ((n-1)^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.

Spoilers ahead!

Suppose n ≠ 0 (modulo 5). Then n^{4} = 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) So n^{4} + 4 is divisible by 5 (and so can only be prime if n = 1).

Suppose on the other hand that n = 0 (modulo 5). If n is even, then so is n^{4}, and so is n^{4} + 4.

So suppose that n is an odd multiple of 5, that is, n = 5(2a + 1) for some a ≥ 0. Then

So n^{4} + 4 cannot be prime in this case either. (For example, 5^{4} + 4 = 17 × 37; and 15^{4} + 4 = 197 × 257.)

I managed the first two steps of this myself; but in fact neither is necessary, there is a perfectly decent proof for all n>1, which your third step comes close to getting.

Ah yes, I see, you can find a similar product for all odd n, not just odd multiples of 5.

Indeed, for all n odd or even (though of course the even ns are trivially proved anyway).

Just to put you out of your misery (if you were in that state):

For n=1, ((n-1)^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.

Spoilers ahead!

Suppose n ≠ 0 (modulo 5). Then n^{4} = 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) So n^{4} + 4 is divisible by 5 (and so can only be prime if n = 1).

Suppose on the other hand that n = 0 (modulo 5). If n is even, then so is n^{4}, and so is n^{4} + 4.

So suppose that n is an odd multiple of 5, that is, n = 5(2a + 1) for some a ≥ 0. Then

So n^{4} + 4 cannot be prime in this case either. (For example, 5^{4} + 4 = 17 × 37; and 15^{4} + 4 = 197 × 257.)

I managed the first two steps of this myself; but in fact neither is necessary, there is a perfectly decent proof for all n>1, which your third step comes close to getting.

Ah yes, I see, you can find a similar product for all odd n, not just odd multiples of 5.

Indeed, for all n odd or even (though of course the even ns are trivially proved anyway).

Just to put you out of your misery (if you were in that state):

For n=1, ((n-1)^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.

Spoilers ahead!

Suppose n ≠ 0 (modulo 5). Then n^{4} = 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) So n^{4} + 4 is divisible by 5 (and so can only be prime if n = 1).

Suppose on the other hand that n = 0 (modulo 5). If n is even, then so is n^{4}, and so is n^{4} + 4.

So suppose that n is an odd multiple of 5, that is, n = 5(2a + 1) for some a ≥ 0. Then

So n^{4} + 4 cannot be prime in this case either. (For example, 5^{4} + 4 = 17 × 37; and 15^{4} + 4 = 197 × 257.)

I managed the first two steps of this myself; but in fact neither is necessary, there is a perfectly decent proof for all n>1, which your third step comes close to getting.

Ah yes, I see, you can find a similar product for all odd n, not just odd multiples of 5.

Indeed, for all n odd or even (though of course the even ns are trivially proved anyway).

Just to put you out of your misery (if you were in that state):

For n=1, ((n-1)^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.

I don’t do formal number proofs, but this one seems obvious: N^4 is N times itself four times, thus it is inherently divisible by itself and four, and adding four to it leaves the resulting number still divisible by four (and an even number, thus divisible by two), at least.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

I don’t do formal number proofs, but this one seems obvious: N^4 is N times itself four times, thus it is inherently divisible by itself and four, and adding four to it leaves the resulting number still divisible by four (and an even number, thus divisible by two), at least.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

I don’t do formal number proofs, but this one seems obvious: N^4 is N times itself four times, thus it is inherently divisible by itself and four, and adding four to it leaves the resulting number still divisible by four (and an even number, thus divisible by two), at least.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

I don’t do formal number proofs, but this one seems obvious: N^4 is N times itself four times, thus it is inherently divisible by itself and four, and adding four to it leaves the resulting number still divisible by four (and an even number, thus divisible by two), at least.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

I don’t do formal number proofs, but this one seems obvious: N^4 is N times itself four times, thus it is inherently divisible by itself and four, and adding four to it leaves the resulting number still divisible by four (and an even number, thus divisible by two), at least.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

I don’t do formal number proofs, but this one seems obvious: N^4 is N times itself four times, thus it is inherently divisible by itself and four, and adding four to it leaves the resulting number still divisible by four (and an even number, thus divisible by two), at least.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

(I spent a week revising solidly for the exam, which was a week after the rest of my exams, got into the exam room and went blank. I left after twenty minutes.)

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

(I spent a week revising solidly for the exam, which was a week after the rest of my exams, got into the exam room and went blank. I left after twenty minutes.)

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

(I spent a week revising solidly for the exam, which was a week after the rest of my exams, got into the exam room and went blank. I left after twenty minutes.)

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

(I spent a week revising solidly for the exam, which was a week after the rest of my exams, got into the exam room and went blank. I left after twenty minutes.)

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

(I spent a week revising solidly for the exam, which was a week after the rest of my exams, got into the exam room and went blank. I left after twenty minutes.)

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

(I spent a week revising solidly for the exam, which was a week after the rest of my exams, got into the exam room and went blank. I left after twenty minutes.)

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

1) (n^4)+4 is always an even number (greater than 2) for all values of n.

2) Even numbers (greater than 2) are never prime.

3) Er… that’s it.

Or am I missing something?

Er, yes. n

^{4}is odd if n is odd; so is n^{4}+4 !!!Oops. Well, I’m only an accountant, not a mathematician.

1) (n^4)+4 is always an even number (greater than 2) for all values of n.

2) Even numbers (greater than 2) are never prime.

3) Er… that’s it.

Or am I missing something?

Er, yes. n

^{4}is odd if n is odd; so is n^{4}+4 !!!Oops. Well, I’m only an accountant, not a mathematician.

1) (n^4)+4 is always an even number (greater than 2) for all values of n.

2) Even numbers (greater than 2) are never prime.

3) Er… that’s it.

Or am I missing something?

Er, yes. n

^{4}is odd if n is odd; so is n^{4}+4 !!!Oops. Well, I’m only an accountant, not a mathematician.

2) Even numbers (greater than 2) are never prime.

3) Er… that’s it.

Or am I missing something?

Er, yes. n

^{4}is odd if n is odd; so is n^{4}+4 !!!Oops. Well, I’m only an accountant, not a mathematician.

2) Even numbers (greater than 2) are never prime.

3) Er… that’s it.

Or am I missing something?

Er, yes. n

^{4}is odd if n is odd; so is n^{4}+4 !!!Oops. Well, I’m only an accountant, not a mathematician.

2) Even numbers (greater than 2) are never prime.

3) Er… that’s it.

Or am I missing something?

Er, yes. n

^{4}is odd if n is odd; so is n^{4}+4 !!!Oops. Well, I’m only an accountant, not a mathematician.

Spoilers ahead!

Suppose

n≠ 0 (modulo 5). Thenn^{4}= 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) Son^{4}+ 4 is divisible by 5 (and so can only be prime ifn= 1).Suppose on the other hand that

n= 0 (modulo 5). Ifnis even, then so isn^{4}, and so isn^{4}+ 4.So suppose that

nis an odd multiple of 5, that is,n= 5(2a+ 1) for somea≥ 0. ThenBut consider the product,

So

n^{4}+ 4 cannot be prime in this case either. (For example, 5^{4}+ 4 = 17 × 37; and 15^{4}+ 4 = 197 × 257.)I managed the first two steps of this myself; but in fact neither is necessary, there is a perfectly decent proof for all n>1, which your third step comes close to getting.

Ah yes, I see, you can find a similar product for all odd

n, not just odd multiples of 5.Indeed, for all

nodd or even (though of course the evenns are trivially proved anyway).Just to put you out of your misery (if you were in that state):

((n-1)

^{2}+1) * ((n+1)^{2}+1)= (n

^{2}-2n +2) * (n^{2}+2n +2)= n

^{4}+ 2n^{3}+ 2n^{2}– 2n^{3}– 4n^{2}– 4n + 2n^{2}+ 4n + 4= n

^{4}+ 4For n=1, ((n-1)

^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.Spoilers ahead!

Suppose

n≠ 0 (modulo 5). Thenn^{4}= 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) Son^{4}+ 4 is divisible by 5 (and so can only be prime ifn= 1).Suppose on the other hand that

n= 0 (modulo 5). Ifnis even, then so isn^{4}, and so isn^{4}+ 4.So suppose that

nis an odd multiple of 5, that is,n= 5(2a+ 1) for somea≥ 0. ThenBut consider the product,

So

n^{4}+ 4 cannot be prime in this case either. (For example, 5^{4}+ 4 = 17 × 37; and 15^{4}+ 4 = 197 × 257.)I managed the first two steps of this myself; but in fact neither is necessary, there is a perfectly decent proof for all n>1, which your third step comes close to getting.

Ah yes, I see, you can find a similar product for all odd

n, not just odd multiples of 5.Indeed, for all

nodd or even (though of course the evenns are trivially proved anyway).Just to put you out of your misery (if you were in that state):

((n-1)

^{2}+1) * ((n+1)^{2}+1)= (n

^{2}-2n +2) * (n^{2}+2n +2)= n

^{4}+ 2n^{3}+ 2n^{2}– 2n^{3}– 4n^{2}– 4n + 2n^{2}+ 4n + 4= n

^{4}+ 4For n=1, ((n-1)

^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.Spoilers ahead!

Suppose

n≠ 0 (modulo 5). Thenn^{4}= 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) Son^{4}+ 4 is divisible by 5 (and so can only be prime ifn= 1).Suppose on the other hand that

n= 0 (modulo 5). Ifnis even, then so isn^{4}, and so isn^{4}+ 4.So suppose that

nis an odd multiple of 5, that is,n= 5(2a+ 1) for somea≥ 0. ThenBut consider the product,

So

n^{4}+ 4 cannot be prime in this case either. (For example, 5^{4}+ 4 = 17 × 37; and 15^{4}+ 4 = 197 × 257.)I managed the first two steps of this myself; but in fact neither is necessary, there is a perfectly decent proof for all n>1, which your third step comes close to getting.

Ah yes, I see, you can find a similar product for all odd

n, not just odd multiples of 5.Indeed, for all

nodd or even (though of course the evenns are trivially proved anyway).Just to put you out of your misery (if you were in that state):

((n-1)

^{2}+1) * ((n+1)^{2}+1)= (n

^{2}-2n +2) * (n^{2}+2n +2)= n

^{4}+ 2n^{3}+ 2n^{2}– 2n^{3}– 4n^{2}– 4n + 2n^{2}+ 4n + 4= n

^{4}+ 4For n=1, ((n-1)

^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.Spoilers ahead!

n≠ 0 (modulo 5). Thenn^{4}= 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) Son^{4}+ 4 is divisible by 5 (and so can only be prime ifn= 1).n= 0 (modulo 5). Ifnis even, then so isn^{4}, and so isn^{4}+ 4.So suppose that

nis an odd multiple of 5, that is,n= 5(2a+ 1) for somea≥ 0. ThenBut consider the product,

n^{4}+ 4 cannot be prime in this case either. (For example, 5^{4}+ 4 = 17 × 37; and 15^{4}+ 4 = 197 × 257.)Ah yes, I see, you can find a similar product for all odd

n, not just odd multiples of 5.Indeed, for all

nodd or even (though of course the evenns are trivially proved anyway).Just to put you out of your misery (if you were in that state):

^{2}+1) * ((n+1)^{2}+1)= (n

^{2}-2n +2) * (n^{2}+2n +2)= n

^{4}+ 2n^{3}+ 2n^{2}– 2n^{3}– 4n^{2}– 4n + 2n^{2}+ 4n + 4= n

^{4}+ 4^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.Spoilers ahead!

n≠ 0 (modulo 5). Thenn^{4}= 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) Son^{4}+ 4 is divisible by 5 (and so can only be prime ifn= 1).n= 0 (modulo 5). Ifnis even, then so isn^{4}, and so isn^{4}+ 4.So suppose that

nis an odd multiple of 5, that is,n= 5(2a+ 1) for somea≥ 0. ThenBut consider the product,

n^{4}+ 4 cannot be prime in this case either. (For example, 5^{4}+ 4 = 17 × 37; and 15^{4}+ 4 = 197 × 257.)Ah yes, I see, you can find a similar product for all odd

n, not just odd multiples of 5.Indeed, for all

nodd or even (though of course the evenns are trivially proved anyway).Just to put you out of your misery (if you were in that state):

^{2}+1) * ((n+1)^{2}+1)= (n

^{2}-2n +2) * (n^{2}+2n +2)= n

^{4}+ 2n^{3}+ 2n^{2}– 2n^{3}– 4n^{2}– 4n + 2n^{2}+ 4n + 4= n

^{4}+ 4^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.Spoilers ahead!

n≠ 0 (modulo 5). Thenn^{4}= 1 (modulo 5). (This is a consequence of Fermat’s little theorem, or you could easily check it by hand.) Son^{4}+ 4 is divisible by 5 (and so can only be prime ifn= 1).n= 0 (modulo 5). Ifnis even, then so isn^{4}, and so isn^{4}+ 4.So suppose that

nis an odd multiple of 5, that is,n= 5(2a+ 1) for somea≥ 0. ThenBut consider the product,

n^{4}+ 4 cannot be prime in this case either. (For example, 5^{4}+ 4 = 17 × 37; and 15^{4}+ 4 = 197 × 257.)Ah yes, I see, you can find a similar product for all odd

n, not just odd multiples of 5.Indeed, for all

nodd or even (though of course the evenns are trivially proved anyway).Just to put you out of your misery (if you were in that state):

^{2}+1) * ((n+1)^{2}+1)= (n

^{2}-2n +2) * (n^{2}+2n +2)= n

^{4}+ 2n^{3}+ 2n^{2}– 2n^{3}– 4n^{2}– 4n + 2n^{2}+ 4n + 4= n

^{4}+ 4^{2}+1) = 1 and so n^{4}+4 is a prime number, but otherwise n^{4}+4 will be composite.I don’t do formal number proofs, but this one seems obvious: N^4 is N times itself four times, thus it is inherently divisible by itself and four, and adding four to it leaves the resulting number still divisible by four (and an even number, thus divisible by two), at least.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

I don’t do formal number proofs, but this one seems obvious: N^4 is N times itself four times, thus it is inherently divisible by itself and four, and adding four to it leaves the resulting number still divisible by four (and an even number, thus divisible by two), at least.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

I don’t do formal number proofs, but this one seems obvious: N^4 is N times itself four times, thus it is inherently divisible by itself and four, and adding four to it leaves the resulting number still divisible by four (and an even number, thus divisible by two), at least.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

Er, no. 3^4 is 81, which is not even; same for any odd number.

Hm, yes, well I did say I didn’t do formal proofs. This rather shows why.

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

(I spent a week revising solidly for the exam, which was a week after the rest of my exams, got into the exam room and went blank. I left after twenty minutes.)

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

(I spent a week revising solidly for the exam, which was a week after the rest of my exams, got into the exam room and went blank. I left after twenty minutes.)

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

(I spent a week revising solidly for the exam, which was a week after the rest of my exams, got into the exam room and went blank. I left after twenty minutes.)

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

Number theory was the only subject that I failed from my undergraduate days, with a stunning 12%

I suspect that strictly speaking this problem is mere algebra rather than yer actual number theory…

A very elementary proof rather.

n^4 + 4 = (n^2 + 2)^2 – 4n^2 = (n^2 + 2 – 2n)(n^2 + 2 + 2n)

As it has 2 factors, it can not be a prime. Hence, proved.

A very elementary proof rather.

n^4 + 4 = (n^2 + 2)^2 – 4n^2 = (n^2 + 2 – 2n)(n^2 + 2 + 2n)

As it has 2 factors, it can not be a prime. Hence, proved.

A very elementary proof rather.

n^4 + 4 = (n^2 + 2)^2 – 4n^2 = (n^2 + 2 – 2n)(n^2 + 2 + 2n)

As it has 2 factors, it can not be a prime. Hence, proved.

A very elementary proof rather.

n^4 + 4 = (n^2 + 2)^2 – 4n^2 = (n^2 + 2 – 2n)(n^2 + 2 + 2n)

As it has 2 factors, it can not be a prime. Hence, proved.

A very elementary proof rather.

n^4 + 4 = (n^2 + 2)^2 – 4n^2 = (n^2 + 2 – 2n)(n^2 + 2 + 2n)

As it has 2 factors, it can not be a prime. Hence, proved.

A very elementary proof rather.

n^4 + 4 = (n^2 + 2)^2 – 4n^2 = (n^2 + 2 – 2n)(n^2 + 2 + 2n)

As it has 2 factors, it can not be a prime. Hence, proved.