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Al Ghailani 1
Proof of a theorem of Fermat that every prime
number of the form 4n + 1 is a sum of two
squares by Leonhard Euler, 1760
Rashid Al Ghailani
Math 464 WI
March 12th, 2024
Al Ghailani 2
Introduction:
1.
When I recently 1 considered these numbers which arise from the
addition of two squares, I proved many properties which such numbers are
endowed with: it was not permitted to prolong adequately my thoughts about this
so that I could show for certain the truth of this theorem, which Fermat once
proposed to be proved via geometry. Nevertheless, I then published an attempt at
the proof, from which the certainty of this theorem shines much brighter, although
it is lacking, according to the criteria of a rigorous proof: I did not doubt that by
continuing in the same path, the desired proof could more easily be obtained,
which indeed since that time came to me with experience, so that the attempt, if
some other slick idea appears, may become a rigorous proof. Indeed, I can boast
of having provided nothing new about this subject, because Fermat himself
already claimed to have elicited a proof of this theorem, but in fact did not make it
public anywhere, just as with the many other exceptional discoveries of this man,
so that what now at last from these lost things we recover, as it were, these things
not unjustly are regarded as new discoveries. Because certainly no one ever so
successfully delved into the arcana of numbers as Fermat, all additional work in
this science to be developed appears to be spent in vain, except previously, which
things from this excellent man are now investigated, as if they are brought to light
anew. Although after him, many learned men in this field of studies have exerted
Al Ghailani 3
their own efforts, still generally nothing has followed which can be compared with
the ingenuity of this man.
2.
As for the proof of the theorem, which I consider here, I have arranged
two propositions it is necessary to call upon for assistance, the proof of which I
have already given elsewhere. The first is that all numbers which are divisors of
sums of two squares prime between themselves are themselves sums of two
squares; thus, if a and b are numbers prime between themselves, and d is a divisor
of a number of the form aa+bb, then d will also be a sum of two squares; I have
given a proof of this theorem in a work mentioned earlier, in which I consider
numbers which are sums of two squares. The first proposition which the following
proof requires is that if p is a prime number and a and b are any numbers not
divisible by p, then ap−1 − bp−1 will always be divisible by the prime number p; I
have already given the proof of this result in Commentarii Academiae
Petropolitanae, Volume VIII.
3.
So if 4n+1 is a prime number, all numbers contained in the form a4n−b4n
will be divisible by it, as long as neither of the numbers a or b is divisible by
4n + 1. Therefore, if a and b are numbers less than 4n + 1 but not equal to 0, then
a number of the form a4n−b4n will be divisible by the given prime number 4n + 1
without any limitation. Because a4n − b4n is a product of the factors a2n + b2n and
a2n − b2n, it is necessary that one of the two factors is divisible by 4n + 1; in other
Al Ghailani 4
words, it cannot be that neither or both have 4n + 1 as a divisor. For if indeed it
can be shown to be the case in which the form a2n+b2n is divisible by 4n+1, since
a2n +b2n, on account of its even exponent 2n, is a sum of two squares, neither of
which is divisible by 4n+1, from this it follows that the number 4n + 1 is a sum of
two squares.
4.
Certainly, the sum a2n + b2n will be divisible by 4n + 1 whenever the
difference a2n − b2n is not divisible by this same number. Therefore, anyone who
will say that the prime number 4n+1 is not a sum of two squares is forced to deny
that any number of the form a2n + b2n is divisible by 4n + 1. Thus, one should affirm
the same, namely, that all numbers contained in the form a2n−b2n are divisible by
4n+1, as long as neither a nor b is divisible by 4n+1. Accordingly, it is to be proved
here by me that not all numbers contained in the form a2n − b2n are divisible by
4n + 1; indeed, in this, if I will have succeeded. it will certainly be the case, if
numbers are substituted for a and b for which the form a2n − b2n is not divisible by
4n + 1; then, in those cases, the other form a2n +b2n will necessarily be divisible by
4n+1. From this, because a2n and b2n are squares, that which is proposed will have
been finished, namely, that the number 4n + 1 is a sum of two squares.
5.
Therefore, so that I may prove that not all numbers contained in the form
a2n − b2n, in other words, not all differences between two powers with exponent
2n, are divisible by 4n + 1, I will consider the series of powers from
1 up to what is formed from the base 4
.
Al Ghailani 5
And now I declare that not all differences between two terms of this series are
divisible by 4n+1. For instance, if the terms of the first difference, 22n−1,32n−
were divisible by 4n + 1, then
the differences of the progression which are the second differences of that series
would also be divisible by 4n + 1; and by the same reasoning, third differences,
fourth, fifth, etc. would all be divisible by 4n + 1, and, finally, also differences of
order 2n, which, as they are constant, are all equal to each other. However, the
differences of order 2n are equal to 1 · 2 · 3 · 4···2n, which therefore are not
divisible by the prime number 4n + 1, from which it follows in turn that certainly
not all first differences are divisible by 4n + 1By virtue of which, the power of this
proof is better observed, it is tobe noted that the difference of order 2n is produced
from 2n + 1 terms of the given series, which, if they are taken from the beginning,
are all so collected that differences of whichever pairs should be divisible by
4n + 1 if the truth of the theorem is negated. But if more terms from this last
constituted difference were assembled, and they were to proceed beyond the term
(4n)2n, seeing that the differences arising from the following term (4n + 1)2n do not
relate to the statement of the theorem, the proof would retain no validity. However,
because the last difference which we are considering depends only on 2n + 1
terms, the conclusion which we deduced from it is entirely legitimate. And from
this it follows that the first differences will be like a2n−(a−1)2n, which is not
divisible by 4n + 1, and in such a way that a is not greater than 2n + 1. However,
Al Ghailani 6
later the sum a2n + (a − 1)2n, is properly obtained, and therefore the sum of two
squares is necessarily divisible by 4n + 1, and thus the prime number 4n + 1 is a
sum of two squares.
6.
Finally, because this proof to be set forth rests on this foundation, that of
the series of powers 1,22n,32n,42n, etc., the differences of order 2n are constant and
all equal 1 · 2 · 3 · 4···2n; this can be seen more fully explained even if it is found
soundly exposited here and there in books of analysis. Therefore, first it is to be
noted that if the general term of any series or, in other words, that which
corresponds to an indefinite exponent of x, is Axm +Bxm−1 +Cxm−2 + Dxm−3 + Exm−4+
etc., then this series is said to be of degree m because m is the exponent of the
highest power of x. Then, if this general term is subtracted from the following
A(x + 1)m + B(x + 1)m−1 + C(x + 1)m−2+ etc., the general term will produce a series of
differences in which the highest power of x will be m−1 and therefore the series
of differences will be of a lower degree, m−1. In a similar way, from the general
term of the series of first differences is collected a general term of the series of
second differences which, again, has an even lower degree, m − 2.
7.
So if the given series is said to be of degree m, the series of first differences
will be of degree m−1, the series of second differences will in turn be of degree
m−2, the series of third differences of degree m−3, the series of fourth differences
will be of degree m − 4, and in general the series of nth order differences will be of
degree m − n. From this, the series of mth order differences will attain degree m−m
= 0, and therefore its general term, because the highest power of x is x0 = 1, will be
Al Ghailani 7
a constant quantity, and therefore all differences of order m will be equal to each
other. Hence, for series of first degree in which the general term is Ax+B, the first
differences will be equal to each other, while for series of second degree, which
are restricted to the general term Ax2 + Bx + C, the second differences are equal,
and so in turn.
8.
If, therefore, we consider any series of powers.
, etc.
for which the general term is xm or, in other words, that which corresponds to a
base of x, the series of differences of order m will be constant from terms equal.
Now the general term of a series of first differences will be.
(x + 1)m − xm,
which, subtracted from the following
(x + 2)m − (x + 1)m
will give the general term of the series of second differences, which will be.
(x + 2)m − 2(x + 1)m + xm.
Thus, in turn, the general term of the series of third differences will be.
(x + 3)m − 3(x + 2)m + 3(x + 1)m − xm;
and finally, the general term of the series of differences of order m is.
+ etc.,
Al Ghailani 8
which, because they are constant quantities, will be the same for whatever number
is substituted for x; therefore, it will be either
+ etc.
or
+ etc.,
where in the former formula we set x = 0 and, in the latter, x = 1.
9.
Let us now consider the special cases of this series and let us ascend
fromthe lowest powers to the highest. And first with m set equal to 1, the general
term of the first differences of the series 1, 2, 3, 4, 5, 6, etc. will be.
either
11 − 1 · 0 1 = 1
or
21 − 1 · 11 = 1.
If m = 2, the second differences of the series 1,22,32,42,52, etc. are
either
22 − 2 · 1 2
or
32 − 2 · 2 2 + 1 · 1 2 ;
but 22 − 2 · 12 = 2(21 − 1 · 11), from which these second differences are 2 · 1.
Let m = 3; the third differences of the series 1,23,33,43,53, etc. will be
either
33 − 3 · 2 3 + 3 · 1 3
or
43 − 3 · 3 3 + 3 · 2 3 − 1 · 1 3 ;
but
33 − 3 · 23 + 3 · 13 = 3(32 − 2 · 22 + 1 · 12) = 3 · 2 · 1,
because from the preceding case,
Al Ghailani 9
32 − 2 · 22 + 1 · 12 = 2 · 1.
In a similar fashion, if m = 4, the fourth differences of the series 1,24,34,44,54, etc.
will be
either
44 − 4 · 3 4 + 6 · 2 4 − 4 · 1 4
or
54 − 4 · 4 4 + 6 · 3 4 − 4 · 2 4 + 1 · 1 4 ;
but
44 − 4 · 34 + 6 · 24 − 4 · 14 = 4(43 − 3 · 33 + 3 · 23 − 1 · 13) = 4 · 3 · 2 · 1.
10. In order that this progression might be more easily observed, let the
differences of order m of the series 1,
, etc. equal P, and let the
differences of order m+1 of the series 1,2m+1,3m+1,4m+1,5m+1, etc. equal Q.
+ etc.,
,
where we expressed P in the former form and Q in the latter form. Here it is evident
that the number of terms is equal in each expression, and individual terms of the
expression P are to the individual terms of the expression Q as 1 is to m + 1. For
instance,
(m + 1)m : (m + 1)m+1 = 1 : m + 1
Al Ghailani 10
On account of this,
P_Q=1:m+1
and thus Q = (m + 1)P.
11. From this it is evident that
for the series
the differences
1, 2, 3, 4, 5, etc.
first = 1,
1,22,32,42,52, etc.
second = 1 · 2
1,23,33,43,53, etc.
third = 1 · 2 · 3
1,24,34,44,54, etc.
fourth= 1 · 2 · 3 · 4
, etc.
of order m = 1 · 2 · 3 · 4 · · · m
and therefore,
, etc.
of order 2n = 1 · 2 · 3 · 4 · · · 2n.
Al Ghailani 11
And thus we have also proved that the differences of order 2n in the series of
powers 1,22n,32n,42n,52n, etc. are not only constant but also equal to the product
1·2·3·4···2n, which we assumed in the proof of the proposed theorem. 3
Conclusion:
Al Ghailani 1
Proof of a theorem of Fermat that every prime
number of the form 4n + 1 is a sum of two
squares by Leonhard Euler, 1760
Rashid Al Ghailani
Math 464 WI
March 12th, 2024
Al Ghailani 2
1. When I recently 1 considered these numbers which arise from the
addition of two squares, I proved many properties which such numbers are
endowed with: it was not permitted to prolong adequately my thoughts about this
so that I could show for certain the truth of this theorem, which Fermat once
proposed to be proved via geometry. Nevertheless, I then published an attempt at
the proof, from which the certainty of this theorem shines much brighter, although
it is lacking, according to the criteria of a rigorous proof: I did not doubt that by
continuing in the same path, the desired proof could more easily be obtained,
which indeed since that time came to me with experience, so that the attempt, if
some other slick idea appears, may become a rigorous proof. Indeed, I can boast
of having provided nothing new about this subject, because Fermat himself
already claimed to have elicited a proof of this theorem, but in fact did not make it
public anywhere, just as with the many other exceptional discoveries of this man,
so that what now at last from these lost things we recover, as it were, these things
not unjustly are regarded as new discoveries. Because certainly no one ever so
successfully delved into the arcana of numbers as Fermat, all additional work in
this science to be developed appears to be spent in vain, except previously, which
things from this excellent man are now investigated, as if they are brought to light
anew. Although after him, many learned men in this field of studies have exerted
their own efforts, still generally nothing has followed which can be compared with
the ingenuity of this man.
2. As for the proof of the theorem, which I consider here, I have arranged
two propositions it is necessary to call upon for assistance, the proof of which I
have already given elsewhere. The first is that all numbers which are divisors of
sums of two squares prime between themselves are themselves sums of two
squares; thus, if a and b are numbers prime between themselves, and d is a divisor
of a number of the form aa+bb, then d will also be a sum of two squares; I have
given a proof of this theorem in a work mentioned earlier, in which I consider
numbers which are sums of two squares. The first proposition which the following
proof requires is that if p is a prime number and a and b are any numbers not
divisible by p, then ap−1 − bp−1 will always be divisible by the prime number p; I
Al Ghailani 3
have already given the proof of this result in Commentarii Academiae
Petropolitanae, Volume VIII.
3. So if 4n+1 is a prime number, all numbers contained in the form a4n−b4n
will be divisible by it, as long as neither of the numbers a or b is divisible by 4n +
1. Therefore, if a and b are numbers less than 4n + 1 but not equal to 0, then a
number of the form a4n−b4n will be divisible by the given prime number 4n + 1
without any limitation. Because a4n − b4n is a product of the factors a2n + b2n and a2n
− b2n, it is necessary that one of the two factors is divisible by 4n + 1; in other
words, it cannot be that neither or both have 4n + 1 as a divisor. For if indeed it
can be shown to be the case in which the form a2n+b2n is divisible by 4n+1, since
a2n +b2n, on account of its even exponent 2n, is a sum of two squares, neither of
which is divisible by 4n+1, from this it follows that the number 4n + 1 is a sum of
two squares.
4. Certainly, the sum a2n + b2n will be divisible by 4n + 1 whenever the
difference a2n − b2n is not divisible by this same number. Therefore, anyone who
will say that the prime number 4n+1 is not a sum of two squares is forced to deny
that any number of the form a2n + b2n is divisible by 4n + 1. Thus, one should affirm
the same, namely, that all numbers contained in the form a2n−b2n are divisible by
4n+1, as long as neither a nor b is divisible by 4n+1. Accordingly, it is to be proved
here by me that not all numbers contained in the form a2n − b2n are divisible by 4n
+ 1; indeed, in this, if I will have succeeded. it will certainly be the case, if numbers
are substituted for a and b for which the form a2n − b2n is not divisible by 4n + 1;
then, in those cases, the other form a2n +b2n will necessarily be divisible by 4n+1.
From this, because a2n and b2n are squares, that which is proposed will have been
finished, namely, that the number 4n + 1 is a sum of two squares.
5. Therefore, so that I may prove that not all numbers contained in the form
a2n − b2n, in other words, not all differences between two powers with exponent
2n, are divisible by 4n + 1, I will consider the series of powers from
1 up to what is formed from the base 4
.
And now I declare that not all differences between two terms of this series are
divisible by 4n+1. For instance, if the terms of the first difference, 22n−1,32n−
were divisible by 4n + 1, then
the differences of the progression which are the second differences of that series
would also be divisible by 4n + 1; and by the same reasoning, third differences,
fourth, fifth, etc. would all be divisible by 4n + 1, and, finally, also differences of
order 2n, which, as they are constant, are all equal to each other. However, the
differences of order 2n are equal to 1 · 2 · 3 · 4···2n, which therefore are not
Al Ghailani 4
divisible by the prime number 4n + 1, from which it follows in turn that certainly
not all first differences are divisible by 4n + 1By virtue of which, the power of this
proof is better observed, it is tobe noted that the difference of order 2n is produced
from 2n + 1 terms of the given series, which, if they are taken from the beginning,
are all so collected that differences of whichever pairs should be divisible by 4n +
1 if the truth of the theorem is negated. But if more terms from this last constituted
difference were assembled, and they were to proceed beyond the term (4n) 2 n,
seeing that the differences arising from the following term (4n + 1)2n do not relate
to the statement of the theorem, the proof would retain no validity. However,
because the last difference which we are considering depends only on 2n + 1
terms, the conclusion which we deduced from it is entirely legitimate. And from
this it follows that the first differences will be like a2n−(a−1)2n, which is not
divisible by 4n + 1, and in such a way that a is not greater than 2n + 1. However,
later the sum a2n + (a − 1)2n, is properly obtained, and therefore the sum of two
squares is necessarily divisible by 4n + 1, and thus the prime number 4n + 1 is a
sum of two squares
6. Finally, because this proof to be set forth rests on this foundation, that of
the series of powers 1,22n,32n,42n, etc., the differences of order 2n are constant and
all equal 1 · 2 · 3 · 4···2n; this can be seen more fully explained even if it is found
soundly exposited here and there in books of analysis. Therefore, first it is to be
noted that if the general term of any series or, in other words, that which
corresponds to an indefinite exponent of x, is Axm +Bxm−1 +Cxm−2 + Dxm−3 + Exm−4+
etc., then this series is said to be of degree m because m is the exponent of the
highest power of x. Then, if this general term is subtracted from the following A(x
+ 1)m + B(x + 1)m−1 + C(x + 1)m−2+ etc., the general term will produce a series of
differences in which the highest power of x will be m−1 and therefore the series
of differences will be of a lower degree, m−1. In a similar way, from the general
term of the series of first differences is collected a general term of the series of
second differences which, again, has an even lower degree, m − 2.
7. So if the given series is said to be of degree m, the series of first differences
will be of degree m−1, the series of second differences will in turn be of degree
m−2, the series of third differences of degree m−3, the series of fourth differences
will be of degree m − 4, and in general the series of nth order differences will be of
degree m − n. From this, the series of mth order differences will attain degree m−m
= 0, and therefore its general term, because the highest power of x is x0 = 1, will be
a constant quantity, and therefore all differences of order m will be equal to each
Al Ghailani 5
other. Hence, for series of first degree in which the general term is Ax+B, the first
differences will be equal to each other, while for series of second degree, which
are restricted to the general term Ax2 + Bx + C, the second differences are equal,
and so in turn.
8.
If, therefore, we consider any series of powers.
, etc.
for which the general term is xm or, in other words, that which corresponds to a
base of x, the series of differences of order m will be constant from terms equal.
Now the general term of a series of first differences will be.
(x + 1)m − xm,
which, subtracted from the following
(x + 2)m − (x + 1)m
will give the general term of the series of second differences, which will be.
(x + 2)m − 2(x + 1)m + xm.
Thus, in turn, the general term of the series of third differences will be.
(x + 3)m − 3(x + 2)m + 3(x + 1)m − xm;
and finally, the general term of the series of differences of order m is.
+ etc.,
which, because they are constant quantities, will be the same for whatever number
is substituted for x; therefore, it will be either
+ etc.
or
+ etc.,
where in the former formula we set x = 0 and, in the latter, x = 1.
9. Let us now consider the special cases of this series and let us ascend
fromthe lowest powers to the highest. And first with m set equal to 1, the general
term of the first differences of the series 1, 2, 3, 4, 5, 6, etc. will be.
either
11 − 1 · 0 1 = 1
or
21 − 1 · 11 = 1.
Al Ghailani 6
If m = 2, the second differences of the series 1,22,32,42,52, etc. are
either
22 − 2 · 1 2
or
32 − 2 · 2 2 + 1 · 1 2 ;
but 22 − 2 · 12 = 2(21 − 1 · 11), from which these second differences are 2 · 1.
Let m = 3; the third differences of the series 1,23,33,43,53, etc. will be
either
33 − 3 · 2 3 + 3 · 1 3
or
43 − 3 · 3 3 + 3 · 2 3 − 1 · 1 3 ;
but
33 − 3 · 23 + 3 · 13 = 3(32 − 2 · 22 + 1 · 12) = 3 · 2 · 1,
because from the preceding case,
32 − 2 · 22 + 1 · 12 = 2 · 1.
In a similar fashion, if m = 4, the fourth differences of the series 1,24,34,44,54, etc.
will be
either
44 − 4 · 3 4 + 6 · 2 4 − 4 · 1 4
or
54 − 4 · 4 4 + 6 · 3 4 − 4 · 2 4 + 1 · 1 4 ;
but
44 − 4 · 34 + 6 · 24 − 4 · 14 = 4(43 − 3 · 33 + 3 · 23 − 1 · 13) = 4 · 3 · 2 · 1.
10. In order that this progression might be more easily observed, let the
differences of order m of the series 1,
, etc. equal P, and let the
differences of order m+1 of the series 1,2m+1,3m+1,4m+1,5m+1, etc. equal Q.
+ etc.,
,
where we expressed P in the former form and Q in the latter form. Here it is evident
that the number of terms is equal in each expression, and individual terms of the
expression P are to the individual terms of the expression Q as 1 is to m + 1. For
instance,
(m + 1)m : (m + 1)m+1 = 1 : m + 1
Al Ghailani 7
On account of this,
P_Q=1:m+1
and thus Q = (m + 1)P.
11. From this it is evident that
for the series
the differences
1, 2, 3, 4, 5, etc.
first = 1,
1,22,32,42,52, etc.
second = 1 · 2
1,23,33,43,53, etc.
third = 1 · 2 · 3
1,24,34,44,54, etc.
fourth= 1 · 2 · 3 · 4
, etc.
of order m = 1 · 2 · 3 · 4 · · · m
and therefore,
, etc.
of order 2n = 1 · 2 · 3 · 4 · · · 2n.
And thus we have also proved that the differences of order 2n in the series of
powers 1,22n,32n,42n,52n, etc. are not only constant but also equal to the product
1·2·3·4···2n, which we assumed in the proof of the proposed theorem. 3
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