Prove that $(5m+3)(3m+1)=n^2$ is not satisfied by any positive integers $m,n$.

I have been staring at this for some time (it's the difficult part of a competition problem, I won't spoil it by naming the problem). I tried looking at it modulo 3,4,5,7,8,16 for a contradiction, as well as looking at the discriminant with respect to $m$. I couldn't finish either way. I wonder whether it can somehow be used that $(-1,2)$ is a solution.

There is a very small chance that there are actually positive solutions but I've ruled out the first few millions for $m$ with a little Python script and, as I said, this was set in a competition so presumably any solutions would be reasonably small.

I've heard somewhere that all diophantine quadratics can be solved by some method and would be interested to read more (though not an entire book, if possible).

  • $\begingroup$ $$3(5m+3)-5(3m+1)=4$$ $\endgroup$ – lab bhattacharjee Feb 15 '18 at 10:37
  • $\begingroup$ Is there perhaps a simple impossibility proof by rewriting the equation as $(4m+2)^2 - (m+1)^2 = n^2$ and using properties of Pythagorean triples? $\endgroup$ – Adam Bailey Feb 16 '18 at 22:29
  • $\begingroup$ Yes, it turns out there is a Pythagorean-style argument using the first comment. The only thing I struggled with there was the case distinction on multiplicities of the common factor $2$ in $(5m+3)$ and $(3m+1)$. It seemed a little lengthy but actually $m=2k+1$ and then $k=2s-1$ for the odd case does it. Thanks for the comments! $\endgroup$ – Max Freiburghaus Feb 18 '18 at 12:18

You can rewrite that equation as $(15m+7)^2-15n^2=4$. Looking at this $\pmod 4$, we can find both $15m+7$ and $n$ need to be even. Hence let $15m+7=2a, \,n=2b$, so that we have the more familiar Pell equation $a^2-15b^2=1$ to solve.

Note the smallest positive solution to the Pell equation is $(a_1, b_1) = (4, 1)$, and all others are generated by $a_n+\sqrt{15}\,b_n = (4+\sqrt{15})^n$, or more particularly of our interest, as a recursion, $a_{n+1} = 8a_n-a_{n-1}$ with $a_1=4, a_2=31$.

Further, note we need $2a=15m+7 \implies 2a \equiv 7 \pmod {15} \implies a \equiv 11 \pmod {15}$. However, the Pell recursion solution above seen $\mod {15}$ generates the sequence $a_i = 4, 1, 4, 1, 4, 1...$ which repeats without ever containing an $11$. Hence there can be no positive solution for $(m, n)$.

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1. Solving $(5m+3)(3m+1)=n^2$
Here is an elementary approach that only uses modular arithmetic.
We do not assuming $m\geq 0$ initially, to highlight where it is used exactly in this approach.

It was shown above that since $$3(5m+3)-5(3m+1)=4$$ We have $d=\gcd(5m+3,3m+1)$ divides $4$.

If $d=1$, then $|5m+3|$ is a square so either $5m+3$ or $-(5m+3)$ is a square. However since squares are $0,1$ or $4$ modulo $5$, this is impossible. Hence we must have $d=2$ or $4$. For either case $m$ is odd so $m=2k+1$.

Now the equation reduces to $$ (10k+8)(6k+4)=n^2 $$ We divide by $4$ into $$ (5k+4)(3k+2) = (n/2)^2 $$ so now $d'=\gcd(5k+4,3k+2)=1$ or $2$.

Suppose first that $d'=2$, so $k$ must be even, say $k=2s$. The new equation after dividing by $4$ is $$ (5s+2)(3s+1) = (n/4)^2 $$ Now $\gcd(5s+2,3s+1)=1$ so, like earlier, $5s+2$ or $-(5s+2)$ must be a square. However this is impossible as squares are $0,1$ or $4$ modulo $5$.

Therefore $d'=1$, hence we have $|3k+2|$ is a square so either $3k+2$ or $-(3k+2)$ is a square. Squares are $0$ or $1$ modulo $3$ so this is impossible for the former. For the latter case, $k\leq -1$ or else $-(3k+2)$ is negative and cannot be square.

We are reduced to a necessary condition that $m=2k+1$ and $k\leq -1$. In particular $k=-1$ is possible, giving $(m,n)=(-1,2)$. This is the part where requiring $m$ to be positive comes into play: As $m=2k+1$, positive $m$ forces $k\geq 0$ and so there are no solutions.

2. Quadratic diophantine in two variables
For two-variable quadratic diophantine equations, of the form $$ax^2+bxy+cy^2+dx+ey+f=0$$ You can refer to posts like this and this.

The idea is you can multiply by integers and use integral linear transformations to convert into $$ X^2 - uY^2 = v $$ so original integer solutions $(x,y)$ transforms into integer solutions $(X,Y)$. By finding all integer solutions to the latter, we can check which ones satisfies the original. The only difficult case is when $u>0$ and not a square, which leads to a series of Pell type equations.

In a more algebraic setting, these questions are well answered by the theory of prime norms in quadratic number fields. Books like Cox's "Primes of form $x^2-ny^2$" addresses it.

For more than two variables indeed there is an algorithm, but unfortunately I do not know much about it.

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  • $\begingroup$ Thanks for the general quadratic! $\endgroup$ – Max Freiburghaus Feb 22 '18 at 9:04
  • $\begingroup$ @MaxFreiburghaus No problem, let me know if I can improve the reference (if it's not clear). $\endgroup$ – Yong Hao Ng Feb 23 '18 at 6:33

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