Navigation   Home   Search   Site map

# document.write (document.title)

 Contact Graeme   Home   Email   Twitter
 Math Help > Number Theory > Squares > 4 Squares in Sequence > 4 Squares in Sequence, pg 4

# Arithmetic Sequence of Four Squares

## Proof 2: No Four Squares In Arithmetic Progression

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 p1², p2², p3² and p4² form a strictly increasing arithmetic progression so that p2² + p3² is minimal over all such progressions.  We will get a contradiction by constructing another progression with smaller middle sum.

We can assume all pi ≥ 0 and, by minimality, the pi have no common factor; hence none has a common factor with the successive difference; hence no successive pair has a common factor.  In particular p1 and p2 cannot both be even.  Considering p1² + p3² = 2 p2² (mod 4), p1 and p2 cannot be odd and even in either order.  (If p1 is even and p2 is odd, then 0 + p3² = 2 (mod 4), and If p1 is odd and p2 is even, then 1 + p3² = 0 ( mod 4); impossible either way).  Hence they are both odd; hence all the pi are odd; hence, ((p-p1) /2)² + ((p3+p1)/2)² = p2² is a Pythagorean triple of squares with no common factor.  By a standard result, there exist opposite-parity, relatively prime a > b > 0 with p2 = a² + b² and (p-p1) /2 and (p3+p1) /2 = 2ab and a² - b² in some order.  Hence p1 = |a² - 2ab - b²|  and p3 = a² + 2ab - b².

Similarly, we can find relatively prime c > d > 0 with p2 = |c² - 2cd - d²|,  p3 = c² + d² and p4 = c² + 2cd - d².

Now we have two representations for p2 (one in terms of a,b and the other in terms of c,d), and two representations for p3.

 p2 = a� + b� p2 = |c� - 2cd - d�| p3 = a� + 2ab - b� p3 = c� + d�

We will equate the two representations of each of the middle elements p2 and p3, then find their sum and difference.  But first, note that one of p2's representations is an absolute value, so we must consider both cases.  Case 1 will be when p2 = c² - 2cd - d², and case 2 will be when p2 = -(c² - 2cd - d²).

Consider Case 1: p2 = c² - 2cd - d².  Equating the two representations for p2 and the two for p3 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 b2=-2cd-d2, 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 2s-t=3B² and 2t-s=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=2A²-C², t=2C²-A².  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 = p2 + p3 < p2² + p3² and we have our contradiction.

Consider Case 2: p2=-(c²-2cd-d²).  As before, equating the two representations for p2 and p3 and taking the sum and difference of the results gives a² + ab = d² + cd and ab-b² = c²-cd.  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(2a²-d²) = c²(2d²-a²).  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: Re-read "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!

### Related Pages in this website

Arithmetic Sequence of Squares -- a series of rambling discussions about the possible existence of an arithmetic sequence of four squares

The webmaster and author of this Math Help site is Graeme McRae.