Prove that no integer solution exists for x⁴ + y⁴ = z² for x,y,z <> 0

We prove using the famous infinite descend method from Fermat

We assume that a smallest integer solution x₁ , y₁ , z₁ does exist.

Therefore (x₁,y₁,z₁) = 1 and so (x₁,y₁)=1 , (x₁,z₁)=1 , (y₁,z₁) = 1

Because (x₁,y₁)=1 both x₁ and y₁ cannot be even. So either both x₁ and y₁ is uneven or x₁ is odd and y₁ is even.

Lets consider the case for both x₁ and y₁ uneven.

Now if x₁ uneven then x₁ = 2n+1 , so x₁⁴ = 4k + 1 and y₁ = 2m+1 so that y₁⁴ = 4h+1

So then z₁² = 4g + 2 = 2(2g+1) = 2h where h is uneven

So z₁² is then even and so must be z₁. But z₁ is the square root of 2h which is clearly irrational

So x₁ must be odd and y₁ must be even and so z₁ must be odd > 0

Therefore y₁⁴ = (z₁ - x₁²)(z₁ + x₁²) = AB --- parx

Where z₁-x₁² = A and z₁ + x₁² = B

Let e | r means e divides r

Because y₁⁴ is even 4 | y₁⁴ and some odd prime t | y₁⁴

Now let us suppose that 4 divides both A and B

Let == suffice for congruent to , so that r == s mod t means t | (r-s)

We have then z₁ == x₁² mod 4 and we have that z₁ == -x₁² mod 4

So 2z₁ == 0 mod 4 and 2x₁² == 0 mod 4

But this cannot be as (x₁,z₁) = 1

The same argument follows for t | y₁⁴

We again get that 2z₁ == 0 mod t and 2x₁² == 0 mod t

But this cannot be as (x₁,z₁) = 1

So we can say that because of the above and because A and B is even that

(A,B) = 2 which means that [(z₁ - x₁²) , (z₁ + x₁²)] = 2 end parx

So we can write y₁ = 2ab where (a,b)=1 and a is odd > 0 and b may be odd or even.

However because y₁ is even we know that at least 16 | y₁⁴ , so for A and B one must have 2 as factor and the other one must have 8 as factor.

So we have either that

Case 1: A = 2a⁴ and B = 8b⁴

or

Case2: that A = 8b⁴ and B = 2a⁴

Now lets consider the first case

Here we have that z₁-x₁² = 2a⁴ and z₁+x₁² = 8b⁴ , so x₁² = -a⁴ + 4b⁴

But x₁ is odd , so we have that x₁ == (1 or 3)mod4

So x₁² == (1 or 9)mod4 , so x₁² == 1 mod4

We see that 4b⁴==0mod4 and we have a odd so that a==(1 or 3)mod4

So a⁴ == (1 or 81)mod4 , so a⁴==(1 or 1)mod4 , so a⁴==1mod4

So -a⁴==-1mod4.

Therefore 4b⁴ - a⁴ == -1mod4 but x₁² = 4b⁴ - a⁴

So x₁² ==-1mod4. We therefore have a contradiction as x₁² cannot both be congruent to 1 and -1 mod4

Case two must therefore be correct and we have that z₁-x₁² = 8b⁴ and z₁+x₁² = 2a⁴

So 2z₁ = 8b⁴ + 2a⁴ . Therefore z₁ = 4b⁴ + a⁴ with a odd > 0 and (a,b)=1 and a<z₁

We also have that 4b⁴ = (a² - x₁)(a²+x₁) , now (a,b)=1 so that (a,x₁)=1

Now going through the same arguments as in parx we get that [(a²-x₁) , (a²+x₁)] = 2

So we can write a² - x₁ = 2x₂⁴ and a² + x₁ = 2y₂⁴

So a⁴ - x₁² = 4x₂⁴y₂⁴ = 4b⁴

So b = x₂y₂

Let a=z₂ as a is odd and so will z₂ be

Now z₂² - x₁ = 2x₂⁴ and z₂² + x₁ = 2y₂⁴

So z² = x₂⁴ + y₂⁴ , but as a<z₁ and thus z₂<z₁ we have a new smaller solution in that the solution (z₁,x₁,y₁) > (z₂,y₂,x₂)

We see we have a contradiction as (z₁,x₁,y₁) was assumed the smallest solution

Therefore there is no integer solution to the equation z² = x⁴ + y⁴

Done.