
The following proof is based very closely on one I found that was written by Joel Karnofsky. It uses the method of " infinite descent".
Start by assuming p_{1}², p_{2}², p_{3}²_{ } and p_{4}² form a strictly increasing arithmetic progression so that p_{2}² +_{ }p_{3}² is minimal over all such progressions. We will get a contradiction by constructing another progression with smaller middle sum.
We can assume all p_{i} ≥ 0 and, by minimality, the p_{i} have no common factor; hence none has a common factor with the successive difference; hence no successive pair has a common factor. In particular p_{1} and_{ }p_{2} cannot both be even. Considering p_{1}² + p_{3}² = 2_{ }p_{2}² (mod 4), p_{1} and_{ }p_{2} cannot be odd and even in either order. (If p_{1} is even and p_{2} is odd, then 0 + p_{3}² = 2 (mod 4), and If p_{1} is odd and p_{2} is even, then 1 + p_{3}² = 0 ( mod 4); impossible either way). Hence they are both odd; hence all the p_{i} are odd; hence, ((p_{3 }p_{1})_{ }/2)² + ((p_{3}+p_{1})/2)² = p_{2}² is a Pythagorean triple of squares with no common factor. By a standard result, there exist oppositeparity, relatively prime a > b > 0 with p_{2} = a² + b² and (p_{3 }p_{1})_{ }/2 and (p_{3}+p_{1})_{ }/2 = 2ab and a²  b² in some order. Hence p_{1} = a²  2ab  b² and p_{3} = a² + 2ab  b².
Similarly, we can find relatively prime c > d > 0 with p_{2} = c²  2cd  d², p_{3} = c² + d² and p_{4} = c² + 2cd  d².
Now we have two representations for p_{2 }(one in terms of a,b and the other in terms of c,d), and two representations for p_{3}.
p_{2} = a� + b� p_{2} = c�  2cd  d� p_{3} = a� + 2ab  b� p_{3} = c� + d�
We will equate the two representations of each of the middle elements p_{2} and p_{3}, then find their sum and difference. But first, note that one of p_{2}'s representations is an absolute value, so we must consider both cases. Case 1 will be when p_{2} = c²  2cd  d², and case 2 will be when p_{2} = (c²  2cd  d²).
Consider Case 1: p_{2} = c²  2cd  d². Equating the two representations for p_{2} and the two for_{ }p_{3} and taking the sum and difference of the results gives a² + ab = c²  cd and ab  b² = cd + d². Note that a ≠ c (if a=c then b^{2}=2cdd^{2}, a negative number). Multiplying the first equation by b² + d² and the second by ab + cd, adding the results and rearranging gives a²(2b² + d²) = c²(b² + 2d²) and in another arrangement b²(2a²  c²) = d²(2c²  a²). Let A and C correspond to a and c after dividing out any common factors (so A ≠ C) and similarly B and D from b and d. Clearly the last two equations can be rewritten with all letters capitalized.
A�(2B� + D�) = C�(B� + 2D�)
B�(2A�  C�) = D�(2C�  A�)
Let s=2B²+D², t=B²+2D². Since 2st=3B² and 2ts=3D², GCD(s,t) = 1 or 3. Checking A²(2B²+D²)=C²(B²+2D²) mod 3, neither B nor D can be divisible by 3 (B and D can't both be divisible by 3, because they have no common factors, so if one is divisible by 3, then A�(1)=C�(1) (mod 3) which implies A and C are both divisible by 3, contradicting A and C having no common factors) so s and t are both multiples of 3 and hence GCD(s,t)=3. Splitting the equation into relatively prime parts gives: A²=(B²+2D²)/3 and (2B²+D²)/3=C².
Now let s=2A²C², t=2C²A². Since 2s+t=3A² and 2t+s=3C², GCD(s,t) = 1 or 3.
In Case 1a, the GCD = 3. Splitting B²(2A²  C²) = D²(2C²  A²) gives B² = (2C²  A²)_{ }/3 and (2A²  C²)_{ }/3 = D². Substituting for B and D in A² = (B² + 2D²)_{ }/3 gives A² = A²/3. Since A is not zero, this case cannot happen.
In Case 1b, the GCD = 1 and splitting gives B² = 2C²  A² and 2A²  C² = D². Rewriting these as B²  C² = C²  A² = A²  D² shows that D², A², C², B² (or this sequence reversed) is a strictly increasing arithmetic sequence of four squares and A² + C² < a² + b² + c² + d² = p_{2} + p_{3} < p_{2}² + p_{3}² and we have our contradiction.
Consider Case 2: p_{2}=(c²2cdd²). As before, equating the two representations for p_{2} and_{ }p_{3} and taking the sum and difference of the results gives a² + ab = d² + cd and abb² = c²cd. Check similary to Case 1 that a ≠ d. Multiplying the first equation by ab + cd and the second by d²  a², adding the results and rearranging gives a²(2b²+c²)=d²(b²+2c²) and in another arrangement b²(2a²d²) = c²(2d²a²). We now proceed as above, with c and d interchanged. As before, Case 2a cannot happen and in Case 2b the sequence C², A², D², B² (or its reverse) gives the contradiction. This completes the proof.
Any proof by contradiction requires you to assume something that is false (although you don't "know" yet that it's false), and then deduce something absurdly false from it. Infinite descent proofs are no different. However, the path from the false assumption to the conclusion of the proof is usually quite involved, and the destination is very specific: unlike other proofs by contradiction, not just any contradiction will do, when you're doing an infinite descent proof. Here, you need to arrive at a particular false statement, namely that a smaller solution satisfies a given set of requirements. In the process, it's easy to fall into the trap of arriving at another contradiction before finding that smaller solution you need.
I'll show you where you can go wrong trying to reproduce the proof on this page: Reread "Case 1", above to understand how we concluded from considering A�(2B�+D�) = C�(B�+2D�) using (mod 3) that (2B�+D�) and (B�+2D�) must share a common factor of 3. Now, take a look at the arguments "Case 1a" and "Case 1b". Here, we studiously avoided telling the reader that the same argument we used in "Case 1" applies here as well, and so (2A�C�) and (2C�A�) must also share 3 as their common factor. If we had done that, we would have arrived at the contradiction mentioned in "Case 1a", which is that A is zero, and we would have never found our smaller solution.
A proof by infinite descent requires shielding the reader from reaching premature contradictions, which isn't always easy. In addition, there's something else you need to do when you're proving by infinite descent: You have to stay on the path of falsehood. Think about it: we're doing a lot of work to derive statements from other statements, every one of which is completely false! As you know, it is possible to validly deduce any statement from a false statement, and so the potential exists that you might accidentally deduce a true statement along the way. If you do, then it will be of no use to you later on, since your objective is to find a smaller solution, which of course doesn't exist. If you stray onto the path of truth during your proof, then you will never "find" that nonexistent smaller solution!
Arithmetic Sequence of Squares  a series of rambling discussions about the possible existence of an arithmetic sequence of four squares
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