# Finding square numbers in arithmetic progression

Say I have three square numbers

(these are squared whole numbers greater than ZERO)

S1 < S2 < S3


how would you go about showing that the following is possible

S2 - S1 = S3 - S2


e.g. the difference between S1 and S2 is the same as that between S2 and S3

• It's not true for 1,4,9 – Akababa Jun 8 '17 at 15:08
• $9-4\ne 4-1{}{}{}$ – peterwhy Jun 8 '17 at 15:09
• I agree, is it true for any set of three squared numbers? – Hector Jun 8 '17 at 15:10
• untrue for all sets of squares – Saketh Malyala Jun 8 '17 at 15:13
• You should think carefully about your question and phrasing it. It often helps you answer your own question. To show it is possible just takes one case. Did you want to know if there are infinite classes of solutions? – Ross Millikan Jun 8 '17 at 15:40

The sequence $1^2,5^2,7^2$ works.
For more solutions you can take $(m^2+2mn-n^2)^2,(m^2+n^2)^2,(n^2+2mn-m^2)^2$ for $n>m$.
Well, start by noticing that the difference between the nth square and the n+1 th square is the nth odd number, if you let the first square be zero. Or, in other words, $$(n+1)^2-n^2=2n+1$$ Which is trivial to prove. Then suppose we have some square numbers $n_1^2, n_2^2, n_3^2$ such that $$n_2^2-n_1^2=n_3^2-n_2^2$$ Let $n_1=n_2-c_1$ and $n_3=n_2+c_2$. Then $$n_2^2-(n_2-c_1)^2=(n_2+c_2)^2-n_2^2$$ $$n_2^2-(n_2^2-2n_2c_1+c_1^2)=(n_2^2+2n_2c_2+c_2^2)-n_2^2$$ $$2n_2c_1-c_1^2=2n_2c_2+c_2^2$$ $$2n_2c_1-2n_2c_2=c_1^2+c_2^2$$ $$n_2^2=\frac{1}{2}\frac{c_1^2+c_2^2}{c_1-c_2}$$ Thus if you find any perfect square $s^2$ in the form $$s^2=\frac{1}{2}\frac{a^2+b^2}{a-b}$$ Then $(s-a)^2, s^2,$ and $(s+b)^2$ should fulfill your requirements.