
The sum of two fourth powers can't be a square, nor can the difference of two fourth powers. Infinite descent proof in 19 cases (patterns) that x^r+y^s=z^t where r,s,t≠2,4.
The first pattern is x^{4}+y^{4} = z^{2}, and if this has no solution, then neither does x^{4}+y^{4} = z^{4}. In other words, we are proving Fermat's last theorem for n = 4. Two fourth powers cannot sum to a fourth power.
We need to consider all these patterns in one go. This is the nature of the proof. The reasoning goes something like this.
If pattern 1 has a solution then there is a smaller triple that satisfies pattern 2. And if pattern 2 has a solution then there is a smaller triple that satisfies pattern 1. Thus every solution implies a smaller solution of the same type, and since positive integers cannot descend forever, there is no solution.
This approach is called the "method of infinite descent", and it is a consequence of induction. If there is a solution there is a least solution, and yet there is always a smaller one, which is a contradiction.
What makes one triple smaller than another? This can be a bit unclear, especially when the triples fit different patterns. The answer is the value of the right hand side. (If the right side is z^{2}, the value is z^{2}, not z.) If one triple has a lesser "value" than another, then it is smaller.
Whenever the coprime lemma applies, we automatically assume coprime variables. After all, this can only make the right side smaller.
x^{2} � u^{2}
 v^{2}
y^{2} � 2uv
z � u^{2} + v^{2}
Since 2uv is a square, and u and v are coprime, u and v are s^{2} and 2t^{2} in some order. Therefore x^{2} becomes s^{4}4t^{4} or 4t^{4}s^{4}. Rewrite this as x^{2}+4t^{4} = s^{4} or x^{2}+s^{4} = 4t^{4}. These are patterns (1) and (2) below.
Each time we defer to another pattern, you should verify that the right side has become smaller. In this case the right side is u^{2} or v^{2}. Either of these is smaller than z, which is smaller than z^{2}, so we're ok.
(�x)^{2} + (�y^{2})^{2} = (z^{2})^{2}
Remember that �x is odd, and so is z^{2}. Characterize the triple as follows.
�x � u^{2}  v^{2}
�y^{2} � 2uv
z^{2} � u^{2}
+ v^{2}
With u and v coprime, the second equation shows u and v are squares. The third equation becomes s^{4}+t^{4} = z^{2}. With z^{2} < 4z^{4}, we have no trouble deferring to pattern (0).
x � u^{2}  v^{2}
2y^{2} � 2uv
z^{2} � u^{2}
+ v^{2}
The second shows u and v are squares, and the third becomes s^{4}+t^{4} = z^{2}. With z^{2} < z^{4}, we defer to pattern (0).
Let either x or z be even, whence the other is also even. If y is even, divide x by 4 and y and z by 2 to find a smaller solution. Thus y is odd. This means x = 2 mod 4, else y would be even. Divide through by 4 to get this.
(�x)^{2} + (y^{2})^{2} = (�z^{2})^{2}
The sum of two odd squares is an even square, and a mod 4 argument shows this is impossible.
Patterns (0), (1), and (2) all fail together, and as a corollary, the sum of fourth powers is never a fourth power. If that's all you needed to know, you can skip the rest of this page, but if you're interested, there are many more patterns with no solutions. Some of these act as lemmas for future theorems.
If y is odd, characterize the triple as y^{2} = u^{2}  v^{2} and z^{2} = u^{2} + v^{2}. Multiply these together to show u^{4}  v^{4} = (yz)^{2}, pattern (3). Note that y < z, hence y^{2}z^{2} < z^{4}.
x^{2} � u^{2}
 v^{2}
2y^{2} � 2uv
z � u^{2} + v^{2}
Thus u and v are squares, and x^{2} = s^{4}t^{4}, pattern (3).
Let x and z be even. If z is divisible by 4 we can divide through and find a smaller solution, so let z = 2 mod 4. This makes y odd. Divide through by 4 and characterize as follows.
y^{2} � u^{2}
 v^{2}
�x^{2} = 2uv
�z = u^{2} + v^{2}
This implies y^{2} = s^{4}t^{4}, pattern (3).
x^{2} �
�(u^{2}
 2v^{2})
y � 2uv
z^{2} � u^{2}
+ 2v^{2}
Multiply the first and third equations together. Now (xz)^{2} is the difference between a fourth power and 4 times a fourth power, patterns (1) and (2).
�x^{2}
�
u^{2}  2uv  v^{2}
y^{2} � u^{2}
+ 2uv  v^{2}
z � u^{2} + v^{2}
Multiply the first two equations together and the left side becomes �(xy)^{2}. Add 8u^{2}v^{2} two both sides and the right side becomes (u^{2}+v^{2})^{2}. This produces the following two equations.
(xy)^{2} + 8u^{2}v^{2} = (u^{2}+v^{2})^{2}
(xy)^{2} + (u^{2}+v^{2})^{2} = 8u^{2}v^{2}
The second equation is impossible mod 4, so concentrate on the first. Write 2st = 2uv, and s^{2}+2t^{2} = u^{2}+v^{2}.
In this characterization, v might be 0, but u cannot be. thus we can replace v with st/u.
s^{2}+2t^{2} = u^{2} + s^{2}t^{2}/u^{2}
(u^{2})^{2}  (s^{2}+2t^{2})u^{2} + s^{2}t^{2} = 0
Apply the quadratic formula. The discriminant is s^{4}+4t^{4}, which must be a square. This never happens, thanks to pattern (4). Of course s or t could be 0; pattern (4) doesn't preclude that. Thus v = 0, and the unique coprime solution emerges: 1^{4} + 1^{4} = 2�1^{2}.
after characterizing the triple, we find u^{2}+v^{2} = 3z^{2}. This is impossible.
If 3 does not divide x and y, there is no solution, thanks to pattern (7), so let 3 divide x and y.
If 3 divides z we can divide through and find a smaller solution, so 3 does not divide z, and x = 3 or 6 mod 9. Let y be even and characterize the triple as follows.
x/3 � u^{2}  v^{2}
y^{2}/3 � 2uv
z^{2} � u^{2}
+ v^{2}
Remember that u and v are coprime, and look at 6uv = y^{2}. Thus u and v are 6s^{2} and t^{2}, or 3s^{2} and 2t^{2}. Substitute in the third equation.
36s^{4} + t^{4} = z^{2}
9s^{4} + 4t^{4} = z^{2}
Since z^{2} is less than 9z^{4}, we defer to patterns (10) and (11).
Next let y be odd and find the following characterization.
y^{2}/3 � u^{2}
 v^{2}
x/3 � 2uv
z^{2} � u^{2}
+ v^{2}
Use the last equation to assign s^{2}t^{2} and 2st to u and v. Then substitute in the first equation.
�y^{2}/3 = s^{4}  6s^{2}t^{2} + t^{4}
The left side can be viewed as �3 times a square. Look at this equation mod 8. Exactly one of s and t is even, hence two of the three terms on the right drop to 0. Now a fourth power is �3 times a square. This is impossible mod 8, so we're done.
Suppose x is divisible by 3. It must be 3 or 6 mod 9, and 3 does not divide y. Derive the following equation, and note that it is impossible mod 3.
(x/3)^{2} + y^{4} = (z^{2}/3)^{2}
Thus the terms are pairwise coprime. There are two different characterizations, depending on whether y is even or odd. In each case z^{2} = u^{2}+v^{2}.
Let y be odd, whence u^{2}v^{2} = 3y^{2}. Multiply this by the equation for z^{2} and find 3(yz)^{2} = u^{4}v^{4}. Since u^{2} < z^{2}, u^{4} < z^{4}, and we defer to pattern (9).
If y is even then 2uv = 3y^{2}. Thus u and v are 6s^{2} and t^{2}, or 3s^{2} and 2t^{2}. Substitute these expressions to set z^{2} equal to 36s^{4} + t^{4} or 4s^{4} + 9t^{4}. These are patterns (10) and (11) respectively.
Let x be even, and z odd, and refer to the characterization of such triples.
�y^{2}
�
u^{2}  3v^{2}
x � 2uv
z^{2} � u^{2}
+ 3v^{2}
Multiply the first and third together to show (yz)^{2} is the difference between a fourth power and 9 times a fourth power. These are patterns (7a) and (8).
Next let z be even, whence x and y are odd. In this case the characterization gives z^{2}/2 = u^{2}+3v^{2}. Since z^{2}/2 is even, u and v have the same parity. This contradicts the characterization, i.e. it makes x and y even as well.
Let 2 divide x and z, while 3 does not. For a minimal solution, z = 2 mod 4, and y is odd. Divide through by 4 and characterize.
3y^{2} � u^{2}
 v^{2}
x^{2}/2 = 2uv
z/2 = u^{2} + v^{2}
The second equation shows u and v are squares, and the first equation sets 3 times a square equal to the difference between fourth powers, which is pattern (9).
Next let 3 divide x and z while 2 does not. For a minimal solution, z is not divisible by 9, and y is not divisible by 3. Divide through by 9 and characterize.
x^{2}/3 � u^{2}
 v^{2}
2y^{2} � 2uv
z/3 � u^{2}+v^{2}
Once again u and v are squares, and the first equation leads to pattern (9).
Finally let 6 divide x and z, but not y, so that we may have a minimal solution.
y^{2} � u^{2}
 v^{2}
x^{2}/6 � 2uv
z/6 � u^{2} + v^{2}
This time u and v are a square and 3 times a square, and the first equation sets x^{2} to the difference between a fourth power and 9 times a fourth power, patterns (7a) and (8).
Let 2 divide y, while 3 does not divide x. Thus 3y^{2}/2 = 2uv, and x^{2} = u^{2}v^{2}. The first equation makes u and v a square and 3 times a square, and the second produces patterns (7a) and (8).
Let 3 divide x, while 2 does not divide y. Thus 2x^{2}/3 = 2uv, and y^{2} = u^{2}v^{2}. The first equation makes u and v a square and 3 times a square, and the second produces patterns (7a) and (8).
finally 3 divides x and 2 divides y. Divide through by 36 for a coprime triple. Thus y^{2}/2 = 2uv, and x^{2}/3 = u^{2}v^{2}. With u and v squares, the second equation leads to pattern (9).
This completes the circle; all patterns up to this point have no solution, save the one solution for pattern (6). But there are more patterns to consider.
If z is even then z^{2} is divisible by 4, which is impossible, since the characterization of z^{2}/2 is an odd number.
x^{4}/4 + 3y^{4} = z^{2}/4
Thus z and x are odd. Characterize this triple and write 2y^{2} = 2uv, and x^{2} = u^{2}3v^{2}. Substitute squares into the second equation to get x^{2} = �(s^{4}3t^{4}). This either sets a sum of squares equal to 3 times a square, which is impossible mod 4, or it creates pattern (12).
Divide through by 4, and only the second term remains even. Characterize, and write y^{2}/2 = 2uv, and x^{2} = u^{2}3v^{2}. Once again x^{2} is the difference between a fourth power and 3 times a fourth power. As we saw in the last pattern, this is impossible, or it creates pattern (12).
The following equation is equivalent.
(z+x^{4}) � (zx^{4}) = 243y^{8}.
If 3 divides both factors it divides x, and we ruled that out. Thus 243 belongs to one of the two factors alone.
Let y be odd, whence one factor is an eighth power, and the other is 243 times an eighth power. Subtract the two equations to get �2x^{4} = u^{8}243v^{8}. Reduce this mod 17. Both u and v cannot drop to 0, as they are coprime. An eighth power is 0 1 or 1. In addition to these, a fourth power could be �4. If u or v is 0 the right side is �1 or �5, while the left side is 0, �2, or �8. This cannot be reconciled, so both u and v are nonzero. The right side becomes �4 or �6. The left side is still 0, �2, or �8. We are forced to conclude that y is even.
�x^{4}
�
u^{2}  3v^{2}
9y^{4} = 2uv
z = u^{2}+3v^{2}
To avoid squares summing to thrice a square, the first equation produces +x^{4}. Thus x^{4}+3v^{2} = u^{2}.
U and v are fourth powers, with coefficients of 8 and 9. If 3 divides u it divides x, hence 9 belongs to v. That leaves two possibilities.
Set u = s^{4} and v = 72t^{4}, giving x^{4} + 15552t^{8} = s^{8}. Thisis pattern (13).
Setting u = 8s^{4} and v = 9t^{4} leads to the following.
x^{4} + 243t^{8} = 64s^{8}
This is impossible mod 8, unless x and t are even. With 64 on the right, x must be divisible by 4. This makes s even, and we can divide through for a smaller solution.
16x^{8} + 243y^{8} = z^{2}
This is impossible mod 4, hence x and z are odd.
If 3 divides x and z then 27 divides z, and 3 divides y, and 81 divides z. We could divide z by 81 and x and y by 3 to find something smaller. Therefore, all terms are coprime.
�x^{4}
�
u^{2}  3v^{2}
36y^{4} � 2uv
z � u^{2}+3v^{2}
This time u and v are fourth powers with coefficients of 2 and 9. If 3 divides u it divides x, hence 9 belongs to v.
Set u = s^{4} and v = 18t^{4}, giving x^{4} + 972t^{8} = s^{8}. This is pattern (13).
Set u = 2s^{4} and v = 9t^{4} and derive x^{4} + 243t^{8} = 4s^{8}. This is pattern (18).
If x and z are divisible by 3, we can make x divisible by 9 and find a smaller solution, as we have done before. Thus the terms are coprime.
The following equation is equivalent.
(2z^{4}+x^{2}) � (2z^{4}x^{2}) = 243y^{8}.
With x and z nonzero mod 3, 243 divides the first factor. Othere than this, both factors are eighth powers.
Add the equations together to get this.
4z^{4} = u^{8}+243v^{8}
With 4z^{4} < 4z^{8}, this is pattern (15).
I want to prove, using mathematical induction, that there are no solutions to the equation x^{4}+y^{4}=z^{4}, for positive values of x, y, z.I assume that x, y, and z are supposed to be integer values; otherwise, there are plenty of solutions.
Those reading this page will likely recognize this question as the n=4 case of Fermat's Last Theorem. The general theorem (that, when n > 2, there are no positive integer solutions to the equation x^{n}+y^{n}=z^{n}), was conjectured by Fermat hundreds of years ago but remained unproved until just recently, when it was proved by Andrew Wiles. The proof is highly complex and involves some deep areas of abstract mathematics.
The n=4 case, however, is relatively easy to prove and was known by Fermat. It turns out to be a little easier to prove the more general result that there is no solution in positive integers to the equation x^{4}+y^{4}=z^{2} (that is, the sum of two fourth powers cannot even be a perfect square, let alone a fourth power).
This is traditionally proven using the "method of infinite descent". The key to this method is the key lemma below (I'll explain later how this fact is proven).
Key Lemma. If x^{4}+y^{4}=z^{2} where x, y, and z are positive integers, then there would exist other positive integers u, v, and w with u^{4}+v^{4}=w^{2}, and w < z.
(For those unfamiliar with the terminology: a lemma is a small theorem whose main use is in proving another, more important, theorem).
The reason this lemma implies that solutions cannot exist is as follows. Suppose solutions did exist. Among all solutions, pick one with the smallest z value (any nonempty set of postive integers has a smallest element). But now the lemma says there would be another solution with a still smaller z value, a contradiction. Therefore, solutions cannot exist.
You can phrase this using the language of mathematical induction, but it's more awkward. You would work by induction on z. When z=1 there are clearly no solutions (x and y have to be at least 1, so z^{2}=x^{4}+y^{4} has to be at least 2).
Now, if it is known that there are no solutions with z≤N, you can prove that there are no solutions with z=N+1 either: if (x,y,z) were such a solution, then the lemma implies that there is another solution (u,v,w) with w < z, so w≤N, contradicting the fact that we know no such solution exists.
The above two paragraphs form the basis and induction steps for a proof by induction. But it's a little cleaner to use the method of infinite descent.
Now for the hard part: a proof of the lemma. I won't fill in all the details (since you originally posted this in the Discussion Topics section rather than the Questions section, I assume you don't want me too), but I will outline the proof.
It depends on the theory of Pythagorean triples. A triple (a,b,c) of positive integers is called a Pythagorean triple if a^{2}+b^{2}=c^{2}. It is called a fundamental Pythagorean triple if a, b, and c have no common factor greater than 1.
For any fundamental Pythagorean triple, one of a and b is even and the other is odd. Let's use the letter a to refer to the even one, b to the odd one. Then there are positive integers m and n such that
a = 2mn
b = m^{2}  n^{2}
c = m^{2} + n^{2}
and m and n are relatively prime (have no common factor greater than 1).
If you don't know how to prove the above statements, feel free to post another question here.
Now, suppose x^{4}+y^{4}=z^{2}. Then (x^{2},y^{2},z) is a Pythagorean triple.
If x, y, and z have a common prime factor p > 1, then (x/p)^{4}+(y/p)^{4}=(z/p^{2})^{2} and we have our desired solution (u,v,w) with u = x/p, v = y/p, and w = z/p^{2} < z. (You need also to explain why z/p^{2} is an integer; I'll leave that to you).
On the other hand, if x, y, and z do not have a common prime factor, then (x^{2},y^{2},z) is a fundamental Pythagorean triple. One of x and y is even and the other odd; let's use x to denote the even one. Then, by the theory of Pythagorean triples, there are relatively prime positive integers m and n such that
x^{2} = 2mn
y^{2} = m^{2}  n^{2}
z = m^{2} + n^{2}
We can rewrite the second equation as n^{2}+y^{2}=m^{2}. We know that m and n have no common factor greater than 1, so (n,y,m) is a fundamental Pythagorean triple. We know y is odd, so n must be the even one. Therefore, appealing to the theory of Pythagorean triples once again, there are relatively prime positive integers r and s such that
n = 2rs
y = r^{2}  s^{2}
m = r^{2} + s^{2}.
The final piece of the puzzle is the fact that if the product of two relatively prime positive integers is a perfect square, then each individually is a perfect square. From this it follows that m and n/2 are perfect squares, since the product is (m)(n/2) = (2mn)/4 = (x^{2})/4 = (x/2)^{2} (remember that x and n are even, so x/2 and n/2 are integers).
It also follows that r and s are perfect squares, since the product is rs = (2rs)/2 = n/2 which is a perfect square.
Therefore, setting r=u^{2}, s=v^{2}, m=w^{2}, the equation m=r^{2}+s^{2} becomes u^{4}+v^{4}=w^{2}. All that remains to complete the proof of the lemma is to show that w < z, which follows because z=m^{2}+n^{2}=w^{4}+n^{2}>w^{4}, so w^{4}<z, and as w is a positive integer this implies w < z as well.
If you want proofs of some of the statements I've left for you to fill in, just post a message here and we'd be glad to fill them in for you.
by Ng Kwong Ming
Posted on Sunday, 14 August, 2005  09:48 am:
Proof that x^{4}y^{4}=z^2 has no solutions.
Turning the equation around slightly,
z^{2}+(y^{2})^{2}=(x^{2})^{2}, so z,y^{2},x^{2} forms a Pythagorean triple.
Assume the equation has a solution in integers. Let x,y,z by the solution with the least x. If GCD(x,y)=d > 1, then we can divide both sides of the equation by d^{4} to obtain a smaller solution, which is impossible if no smaller value than x solves the equation. This forces x to be odd, because no primitive Pythagorean triple has an even number in the third position.
y is either even or odd. I will consider both cases:
case 1: y is odd.
y^{2},z,x^{2} forms a primitive Pythagorean triple, with y odd.
z=2st, s>t, GCD(s,t)=1, s,t are of opposite parity.
y^{2}=s^{2}t^{2}
x^{2}=s^{2}+t^{2}
s^{4}t^{4}=(xy)^{2}, so it is a solution to the equation.
0<s<sqrt(s^{2}+t^{2})=x, contradicting minimal value of x
case 2: y is even.
z,y^{2},x^{2} forms a primitive Pythagorean triple, with y even.
y^{2}=2st
z=s^{2}t^{2}
x^{2}=s^{2}+t^{2}.If s is even and t odd, then from y^{2}=2st, we have GCD(2s,t)=1, so 2s=w^{2}, t=v^{2}.
Since w is even, w=2u, and s=2u^{2}
And x^{2}=s^{2}+t^{2}=4u^{4}+v^{4}, so v^{2},2u^{2},x form a primitive Pythagorean triple with
2u^{2}=2ab,
v^{2}=a^{2}b^{2}, and
x=a^{2}+b^{2}.
u^{2}=ab implies a,b are squares, so a=c^{2}, b=d^{2}.
v^{2}=a^{2}b^{2}=c^{4}d^{4}. This results in a new solution c, d, v of x^{4}y^{4}=z^{2}, and what is more,
0<c=sqrt(a)<a^{2}+b^{2}=x, contrary to our assumption regarding x.If t is even and s odd, then from y^{2}=2st, we have GCD(s,2t)=1, so s=w^{2}, 2t=v^{2}.
Since v is even, v=2u, and t=2u^{2}
And x^{2}=s^{2}+t^{2}=w^{4}+4u^{4}, so w^{2},2u^{2},x form a primitive Pythagorean triple, and then the rest of it follows the same logic as the preceding paragraph.
Combining the above, the only resolution of these contradictions is that the equation cannot be satisfied in integers
http://www.math.toronto.edu/mathnet/questionCorner/fermat4.html  proof that x^4+y^4=z^2 has no solution
http://www.mathreference.com/numzext,442.html  proof that eighteen different patterns of equations involving two fourth powers and a square have no solution by a convoluted finite state machine using infinite descent with each transition.
The webmaster and author of this Math Help site is Graeme McRae.